Trigonometrical Ratios

Given:

sin $A = \frac{5}{13}$

i.e., $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13}$

∴ If length of BC = 5x unit, length of AC = 13x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (13x)² = AB² + (5x)²

⇒ 169x² = AB² + 25x²

⇒ AB² = 169x² - 25x²

⇒ AB² = 144x²

⇒ AB = $\sqrt{144x^2}$

⇒ AB = 12x

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{BC}{AB}$ = $\frac{5x}{12x}$ = $\frac{5}{12}$

Hence, option 1 is the correct option.

Given:

tan $A = \frac{3}{5}$

i.e. $\frac{Perpendicular}{Base} = \frac{3}{5}$

∴ If length of BC = 3x unit, length of AB = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (3x)²

⇒ AC² = 25x² + 9x²

⇒ AC² = 34x²

⇒ AC = $\sqrt{34\text{x}^2}$

⇒ AC = $\sqrt{34}$ x

sin $A = \frac{Perpendicular}{Hypotenuse}$ =

$= \frac{BC}{AC} = \frac{3x}{\sqrt{34}x} = \frac{3}{\sqrt{34}}$

cos $A = \frac{Base}{Hypotenuse}$

$= \frac{AB}{AC} = \frac{5x}{\sqrt{34}x} = \frac{5}{\sqrt{34}}$

Now, sin² A + cos² A

$$\begin{aligned}= \Big(\frac{3}{\sqrt{34}}\Big)^2 + \Big(\frac{5}{\sqrt{34}}\Big)^2\\[1em] = \Big(\frac{9}{34}\Big) + \Big(\frac{25}{34}\Big)\\[1em] = \Big(\frac{9 + 25}{34}\Big)\\[1em] = \Big(\frac{34}{34}\Big)\\[1em] = 1\end{aligned}$$

Hence, option 2 is the correct option.

Given:

cot $A = \frac{5}{12}$

i.e. $\frac{Base}{Perpendicular} = \frac{5}{12}$

∴ If length of AB = 5x unit, length of BC = 12x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$\frac{AC}{BC} = \frac{13x}{12x} = \frac{13}{12}$

Now, cot² A - cosec² A

$$\begin{aligned}= \Big(\frac{5}{12}\Big)^2 - \Big(\frac{13}{12}\Big)^2\\[1em] = \Big(\frac{25}{144}\Big) - \Big(\frac{169}{144}\Big)\\[1em] = \Big(\frac{25 - 169}{144}\Big)\\[1em] = \Big(\frac{- 144}{144}\Big)\\[1em] = - 1\end{aligned}$$

Hence, option 4 is the correct option.

In Δ ABD,

AB² = AD² + BD²

⇒ (26)² = AD² + (10)²

⇒ 676 = AD² + 100

⇒ AD² = 676 - 100

⇒ AD² = 576

⇒ AD = $\sqrt{576}$

⇒ AD = 24 cm

In Δ ADC,

AC² = AD² + DC²

⇒ AC² = (24)² + (32)²

⇒ AC² = 576 + 1,024

⇒ AC² = 1,600

⇒ AC = $\sqrt{1,600}$

⇒ AC = 40 cm

tan $A = \frac{Perpendicular}{Base}$

= $\frac{24}{32}$

= $\frac{3}{4}$

Hence, option 3 is the correct option.

Given:

5 cos A = 3

⇒ cos A = $\frac{3}{5}$

i.e., $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

AC² = BC² + AB²

⇒ (5x)² = BC² + (3x)²

⇒ 25x² = BC² + 9x²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

sec A = $\frac{Hypotenuse}{Base}$

= $\frac{AC}{BA} = \frac{5x}{3x} = \frac{5}{3}$

tan A = $\frac{Perpendicular}{Base}$

= $\frac{BC}{BA} = \frac{4x}{3x} = \frac{4}{3}$

Now, sec² A - tan² A

$$\begin{aligned}= \Big(\frac{5}{3}\Big)^2 - \Big(\frac{4}{3}\Big)^2\\[1em] = \frac{25}{9} - \frac{16}{9}\\[1em] = \frac{25 - 16}{9}\\[1em] = \frac{9}{9}\\[1em] = 1\end{aligned}$$

Hence, option 1 is the correct option.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 3² + 4²

⇒ AC² = 9 + 16

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

(i) sin $A = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{CB}{AC}\\[1em] = \frac{4}{5}\\[1em]\end{aligned}$$

Hence, sin $A = \frac{4}{5}$.

(ii) cos $A = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{AB}{AC}\\[1em] = \frac{3}{5}\\[1em]\end{aligned}$$

Hence, cos $A = \frac{3}{5}$.

(iii) cot $A = \frac{Base}{Perpendicular}$

$$\begin{aligned}= \frac{AB}{BC}\\[1em] = \frac{3}{4}\\[1em]\end{aligned}$$

Hence, cot $A = \frac{3}{4}$.

(iv) sec $C = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{AC}{AB}\\[1em] = \frac{5}{4}\\[1em] = 1\frac{1}{4}\\[1em]\end{aligned}$$

Hence, sec $A = \frac{5}{4} = 1\frac{1}{4}$.

(v) cosec $C = \frac{Hypotenuse}{Perpendicular}$

$$\begin{aligned}= \frac{AC}{AB}\\[1em] = \frac{5}{3}\\[1em] = 1\frac{2}{3}\\[1em]\end{aligned}$$

Hence, cosec $C = \frac{5}{3} = 1\frac{2}{3}$.

(vi) tan $C = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{AB}{BC}\\[1em] = \frac{3}{4}\\[1em]\end{aligned}$$

Hence, tan $C = \frac{3}{4}$.

In Δ BAC,

⇒ BC² = AB² + AC² (∵ BC is hypotenuse)

⇒ (17)² = (8)² + AC²

⇒ 289 = 64 + AC²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) cos $B = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{AB}{BC}\\[1em] = \frac{8}{17}\end{aligned}$$

Hence, cos $B = \frac{8}{17}$.

(ii) tan $C = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{AB}{AC}\\[1em] = \frac{8}{15}\\[1em]\end{aligned}$$

Hence, tan $C = \frac{8}{15}$.

(iii) sin²B + cos²B

$$\begin{aligned}= \Big(\frac{Perpendicular}{Hypotenuse}\Big)^2 + \Big(\frac{Base}{Hypotenuse}\Big)^2\\[1em] = \Big(\frac{AC}{BC}\Big)^2 + \Big(\frac{AB}{BC}\Big)^2\\[1em] = \Big(\frac{15}{17}\Big)^2 + \Big(\frac{8}{17}\Big)^2\\[1em] = \frac{225}{289} + \frac{64}{289}\\[1em] = \frac{225 + 64}{289}\\[1em] = \frac{289}{289}\\[1em] = 1\end{aligned}$$

Hence, sin²B + cos²B = 1.

(iv) sin B.cos C + cos B.sin C

$$\begin{aligned}= \frac{Perpendicular}{Hypotenuse} .\frac{Base}{Hypotenuse} + \frac{Base}{Hypotenuse} . \frac{Perpendicular}{Hypotenuse}\\[1em] = \frac{AC}{BC} .\frac{AC}{BC} + \frac{AB}{BC} . \frac{AB}{BC}\\[1em] = \Big(\frac{AC}{BC}\Big)^2 + \Big(\frac{AB}{BC}\Big)^2\\[1em] = \Big(\frac{15}{17}\Big)^2 + \Big(\frac{8}{17}\Big)^2\\[1em] = \frac{225}{289} + \frac{64}{289}\\[1em] = \frac{225 + 64}{289}\\[1em] = \frac{289}{289}\\[1em] = 1\end{aligned}$$

Hence, sin B.cos C + cos B.sin C = 1.

In Δ ADB,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = (3)² + (4)²

⇒ AB² = 9 + 16

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5

In Δ CDB,

⇒ BC² = BD² + DC² (∵ BC is hypotenuse)

⇒ (12)² = (4)² + DC²

⇒ 144 = 16 + DC²

⇒ DC² = 144 - 16

⇒ DC² = 128

⇒ DC = $\sqrt{128}$

⇒ DC = $8\sqrt{2}$

(i) cos $A = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{AD}{AB}\\[1em] = \frac{3}{5}\end{aligned}$$

Hence, cos $A = \frac{3}{5}$.

(ii) cosec $A = \frac{Hypotenuse}{Perpendicular}$

$$\begin{aligned}= \frac{AD}{BD}\\[1em] = \frac{5}{4}\end{aligned}$$

Hence, cosec $A = \frac{5}{4}$.

(iii) tan²A – sec²A

tan $A = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{BD}{AD}\\[1em] = \frac{4}{3}\end{aligned}$$

sec $A = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{AB}{AD}\\[1em] = \frac{5}{3}\end{aligned}$$

tan²A – sec²A

$$\begin{aligned}= \Big(\frac{4}{3}\Big)^2 - \Big(\frac{5}{3}\Big)^2\\[1em] = \Big(\frac{16}{9}\Big) - \Big(\frac{25}{9}\Big)\\[1em] = \Big(\frac{16 - 25}{9}\Big)\\[1em] = \Big(\frac{-9}{9}\Big)\\[1em] = -1\end{aligned}$$

Hence, tan²A – sec²A = -1.

(iv) sin $C = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{BD}{BC}\\[1em] = \frac{4}{12}\\[1em] = \frac{1}{3}\\[1em]\end{aligned}$$

Hence, sin $C = \frac{1}{3}$.

(v) sec $C = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{BC}{DC}\\[1em] = \frac{12}{8 \sqrt{2}}\\[1em] = \frac{3}{2 \sqrt{2}}\\[1em] = \frac{3 \times \sqrt{2}}{2 \sqrt{2} \times \sqrt{2}}\\[1em] = \frac{3 \sqrt{2}}{2 \times 2}\\[1em] = \frac{3 \sqrt{2}}{4}\end{aligned}$$

Hence, sec $C = \frac{3 \sqrt{2}}{4}$.

(vi) cot²C - $\frac{1}{\text{sin}^2 \text{ C}}$

cot $C = \frac{Base}{Perpendicular}$

$$\begin{aligned}= \frac{Base}{Perpendicular}\\[1em] = \frac{8 \sqrt{2}}{4}\\[1em] = 2 \sqrt{2}\end{aligned}$$

sin $C = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{4}{12}\\[1em] = \frac{1}{3}\end{aligned}$$

Now, cot²C - $\frac{1}{\text{sin}^2 \text{ C}}$

$$\begin{aligned}= (2 \sqrt{2})^2 - \frac{1}{\Big(\frac{1}{3}\Big)^2}\\[1em] = (2 \sqrt{2})^2 - \frac{9}{1}\\[1em] = 4 \times 2 - 9\\[1em] = 8 - 9\\[1em] = -1\end{aligned}$$

Hence, cot²C - $\frac{1}{\text{sin}^2 \text{ C}} = -1$.

In Δ ABD,

⇒ AB² = BD² + DA² (∵ AB is hypotenuse)

⇒ 13² = 5² + DA²

⇒ 169 = 25 + DA²

⇒ DA² = 169 - 25

⇒ DA² = 144

⇒ DA = $\sqrt{144}$

⇒ DA = 12

In Δ ADC,

⇒ AC² = AD² + DC² (∵ AB is hypotenuse)

⇒ AC² = 12² + 16²

⇒ AC² = 144 + 256

⇒ AC² = 400

⇒ AC = $\sqrt{400}$

⇒ AC = 20

(i) sin $B = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{AD}{AB}\\[1em] = \frac{12}{13}\end{aligned}$$

Hence, sin $B = \frac{12}{13}$.

(ii) tan $C = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{AD}{DC}\\[1em] = \frac{12}{16}\\[1em] = \frac{3}{4}\end{aligned}$$

Hence, tan $C = \frac{3}{4}$.

(iii) sec² B – tan² B

sec $B = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{AB}{BD}\\[1em] = \frac{13}{5}\end{aligned}$$

tan $B = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{AD}{BD}\\[1em] = \frac{12}{5}\end{aligned}$$

sec² B – tan² B

$$\begin{aligned}= \Big(\frac{13}{5}\Big)^2 - \Big(\frac{12}{5}\Big)^2\\[1em] = \Big(\frac{169}{25}\Big) - \Big(\frac{144}{25}\Big)\\[1em] = \Big(\frac{169 - 144}{25}\Big)\\[1em] = \Big(\frac{25}{25}\Big)\\[1em] = 1\end{aligned}$$

Hence, sec² B – tan² B = 1.

(iv) sin² C + cos² C

sin $C = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{AD}{AC}\\[1em] = \frac{12}{20}\\[1em] = \frac{3}{5}\end{aligned}$$

cos $C = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{DC}{AC}\\[1em] = \frac{16}{20}\\[1em] = \frac{4}{5}\end{aligned}$$

Now,

sin² C + cos² C

$$\begin{aligned}= \Big(\frac{3}{5}\Big)^2 + \Big(\frac{4}{5}\Big)^2\\[1em] = \Big(\frac{9}{25}\Big) + \Big(\frac{16}{25}\Big)\\[1em] = \Big(\frac{9 + 16}{25}\Big)\\[1em] = \Big(\frac{25}{25}\Big)\\[1em] = 1\end{aligned}$$

Hence, sin² C + cos² C = 1.

(i) Given:

sin $A = \frac{3}{5}$

i.e., $\frac{Perpendicular}{Hypotenuse} = \frac{3}{5}$

∴ If length of BC = 3x unit, length of AC = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (5x)² = AB² + (3x)²

⇒ 25x² = AB² + 9x²

⇒ AB² = 25x² - 9x²

⇒ AB² = 16x²

⇒ AB = $\sqrt{16\text{x}^2}$

⇒ AB = 4x

tan $A = \frac{Perpendicular}{Base}$

= $\frac{BC}{BA} = \frac{3x}{4x} = \frac{3}{4}$

Hence, tan $A = \frac{3}{4}$.

(ii) cos $A = \frac{Base}{Hypotenuse}$

$= \frac{BA}{AC} =\frac{4x}{5x} = \frac{4}{5}$

Hence, cos $A = \frac{4}{5}$.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = a² + a²

⇒ AC² = 2a²

⇒ AC = $\sqrt{2a^2}$

⇒ AC = $a \sqrt{2}$

(i) sin $A = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{CB}{AC}\\[1em] = \frac{a}{a \sqrt{2}}\\[1em] = \frac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \frac{a \sqrt{2}}{a \times 2}\\[1em] = \frac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \frac{\sqrt{2}}{2}\\[1em]\end{aligned}$$

Hence, sin $A = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.

(ii) sec $A = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{AC}{AB}\\[1em] = \frac{a \sqrt{2}}{a}\\[1em] = \frac{\cancel{a} \sqrt{2}}{\cancel{a}}\\[1em] = \sqrt{2}\end{aligned}$$

Hence, sec $A = \sqrt{2}$.

(iii) cos² A + sin² A

cos $A = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{AB}{AC}\\[1em] = \frac{a}{a \sqrt{2}}\\[1em] = \frac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \frac{a \sqrt{2}}{a \times 2}\\[1em] = \frac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \frac{\sqrt{2}}{2}\\[1em]\end{aligned}$$

sin $A = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{CB}{AC}\\[1em] = \frac{a}{a \sqrt{2}}\\[1em] = \frac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \frac{a \sqrt{2}}{a \times 2}\\[1em] = \frac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \frac{\sqrt{2}}{2}\\[1em]\end{aligned}$$

Now, cos² A + sin² A

$$\begin{aligned}= \Big(\frac{\sqrt{2}}{2}\Big)^2 + \Big(\frac{\sqrt{2}}{2}\Big)^2\\[1em] = \frac{2}{4} + \frac{2}{4}\\[1em] = \frac{2 + 2}{4}\\[1em] = \frac{4}{4}\\[1em] = 1\end{aligned}$$

Hence, cos² A + sin² A = 1.

Given:

cos $A = \frac{5}{13}$

i.e. $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{5}{13}$

∴ If length of BA = 5x unit, length of AC = 13x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (13x)² = (5x)² + BC²

⇒ 169x² = 25x² + BC²

⇒ BC² = 169x² - 25x²

⇒ BC² = 144x²

⇒ BC = $\sqrt{144\text{x}^2}$

⇒ BC = 12x

(i) sin $A = \frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{AC} = \frac{12x}{13x} = \frac{12}{13}$

cot $A = \frac{Base}{Perpendicular}$

$= \frac{AB}{BC} = \frac{5x}{12x} = \frac{5}{12}$

tan $A = \frac{Perpendicular}{Base}$

$= \frac{BC}{AB} = \frac{12x}{5x} = \frac{12}{5}$

Now,

$$\begin{aligned}= \frac{\text{sin A} - \text{cot A}}{2 \text{ tan A}}\\[1em] = \frac{\frac{12}{13} - \frac{5}{12}}{2 \times \frac{12}{5}}\\[1em] = \frac{\frac{12 \times 12}{13 \times 12} - \frac{5 \times 13}{12 \times 13}}{\frac{24}{5}}\\[1em] = \frac{\frac{144}{156} - \frac{65}{156}}{\frac{24}{5}}\\[1em] = \frac{\frac{144 - 65}{156}}{\frac{24}{5}}\\[1em] = \frac{\frac{79}{156}}{\frac{24}{5}}\\[1em] = \frac{79 \times 5}{156 \times 24}\\[1em] = \frac{395}{3,744}\end{aligned}$$

Hence, $\frac{\text{sin A} - \text{cot A}}{2 \text{ tan A}} = \frac{395}{3,744}$.

(ii) cos $A = \frac{5}{13}$

cot $A = \frac{5}{12}$

To find,

$\text{cot A} + \frac{1}{\text{cos A}}$

$$\begin{aligned}\text{cot A} + \frac{1}{\text{cos A}} = \frac{5}{12} + \frac{1}{\frac{5}{13}}\\[1em] = \frac{5}{12} + \frac{13}{5}\\[1em] = \frac{5 \times 5}{12 \times 5} + \frac{13 \times 12}{5 \times 12}\\[1em] = \frac{25}{60} + \frac{156}{60}\\[1em] = \frac{25 + 156}{60}\\[1em] = \frac{181}{60}\end{aligned}$$

Hence, $\text{cot A} + \frac{1}{\text{cos A}} = \frac{181}{60}$.

Given:

sec $A = \frac{29}{21}$

i.e. $\frac{\text{Hypotenuse}}{\text{Base}} = \frac{29}{21}$

∴ If length of AB = 21x unit, length of AC = 29x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (29x)² = (21x)² + BC²

⇒ 841x² = 441x² + BC²

⇒ BC² = 841x² - 441x²

⇒ BC² = 400x²

⇒ BC = $\sqrt{400\text{x}^2}$

⇒ BC = 20x

sin $A = \frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{AC} = \frac{20x}{29x} = \frac{20}{29}$

tan $A = \frac{Perpendicular}{Base}$

$= \frac{BC}{AB} = \frac{20x}{21x} = \frac{20}{21}$

Now,

$$\begin{aligned}\text{sin A} - \frac{1}{\text{tan A}}\\[1em] = \frac{20}{29} - \frac{1}{\frac{20}{21}}\\[1em] = \frac{20}{29} - \frac{21}{20}\\[1em] = \frac{20 \times 20}{29 \times 20} - \frac{21 \times 29}{20 \times 29}\\[1em] = \frac{400}{580} - \frac{609}{580}\\[1em] = \frac{400 - 609}{580}\\[1em] = \frac{- 209}{580}\\[1em]\end{aligned}$$

Hence, $\text{sin A} - \frac{1}{\text{tan A}} = \frac{- 209}{580}$.

Given:

tan $A = \frac{4}{3}$

i.e., $\frac{\text{Perpendicular}}{\text{Base}} = \frac{4}{3}$

∴ If length of AB = 3x unit, length of BC = 4x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$= \frac{AC}{BC} = \frac{5x}{4x} = \frac{5}{4}$

cot $A = \frac{Base}{Perpendicular}$

$= \frac{AB}{BC} = \frac{3x}{4x} = \frac{3}{4}$

sec $A = \frac{Hypotenuse}{Base}$

$= \frac{AC}{AB} = \frac{5x}{3x} = \frac{5}{3}$

Now,

$$\begin{aligned}\frac{\text{cosec A}}{\text{cot A} - \text{sec A}}\\[1em] = \frac{\frac{5}{4}}{\frac{3}{4} - \frac{5}{3}}\\[1em] = \frac{\frac{5}{4}}{\frac{3 \times 3}{4 \times 3} - \frac{5 \times 4}{3 \times 4}}\\[1em] = \frac{\frac{5}{4}}{\frac{9}{12} - \frac{20}{12}}\\[1em] = \frac{\frac{5}{4}}{\frac{9 - 20}{12}}\\[1em] = \frac{\frac{5}{4}}{\frac{- 11}{12}}\\[1em] = \frac{5 \times 12}{- 11 \times 4}\\[1em] = \frac{- 60}{44}\\[1em] = \frac{- 15}{11}\end{aligned}$$

Hence, $\frac{\text{cosec A}}{\text{cot A} - \text{sec A}} = \frac{- 15}{11}$.

Given:

4 cot A = 3

cot $A = \frac{3}{4}$

i.e., $\frac{\text{Base}}{\text{Perpendicular}} = \frac{3}{4}$

∴ If length of AB = 3x unit, length of BC = 4x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

(i) sin $A = \frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{AC} = \frac{4x}{5x} = \frac{4}{5}$

Hence, sin $A = \frac{4}{5}$.

(ii) sec $A = \frac{Hypotenuse}{Base}$

$= \frac{AC}{AB} = \frac{5x}{3x} = \frac{5}{3} = 1\frac{2}{3}$

Hence, sec $A = \frac{5}{3} = 1\frac{2}{3}$.

(iii) cosec² A - cot² A

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$= \frac{AC}{BC} = \frac{5x}{4x} = \frac{5}{4}$

cot $A = \frac{Base}{Perpendicular}$

$= \frac{AB}{BC} = \frac{3x}{4x} = \frac{3}{4}$

Now,

$$\begin{aligned}\text{cosec}^2 A - \text{cot}^2 A\\[1em] = \Big(\frac{5}{4}\Big)^2 - \Big(\frac{3}{4}\Big)^2\\[1em] = \frac{25}{16} - \frac{9}{16}\\[1em] = \frac{25 - 9}{16}\\[1em] = \frac{16}{16}\\[1em] = 1\end{aligned}$$

Hence, cosec² A - cot² A = 1.

Given:

cos A = 0.6

cos $A = \frac{6}{10}$

cos $A = \frac{3}{5}$

i.e. $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (5x)² = (3x)² + BC²

⇒ 25x² = 9x² + BC²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

sin $A = \frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{AC} = \frac{4x}{5x} = \frac{4}{5}$

tan $A = \frac{Perpendicular}{Base}$

$= \frac{BC}{AB} = \frac{4x}{3x} = \frac{4}{3} = 1\frac{1}{3}$

cot $A = \frac{Base}{Perpendicular}$

$= \frac{AB}{BC} = \frac{3x}{4x} = \frac{3}{4}$

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$= \frac{AC}{BC} = \frac{5x}{4x} = \frac{5}{4} = 1\frac{1}{4}$

sec $A = \frac{Hypotenuse}{Base}$

$= \frac{AC}{AB} = \frac{5x}{3x} = \frac{5}{3} = 1\frac{2}{3}$

Hence, sin $A = \frac{4}{5}$, tan $A = 1\frac{1}{3}$, cot $A = \frac{3}{4}$, cosec $A = 1\frac{1}{4}$ and sec $A = 1\frac{2}{3}$.

Given:

tan $A = \frac{5}{12}$

i.e., $\frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12}$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

(i) cos $A = \frac{Base}{Hypotenuse}$

$= \frac{AB}{AC} = \frac{12x}{13x} = \frac{12}{13}$

Hence, cos $A = \frac{12}{13}$.

(ii) sin $A = \frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{AC} = \frac{5x}{13x} = \frac{5}{13}$

Hence, sin $A = \frac{5}{13}$.

(iii) $\frac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}}$

$$\begin{aligned}= \frac{\frac{12}{13} + \frac{5}{13}}{\frac{12}{13} - \frac{5}{13}}\\[1em] = \frac{\frac{12 + 5}{13}}{\frac{12 - 5}{13}}\\[1em] = \frac{\frac{17}{\cancel{13}}}{\frac{7}{\cancel{13}}}\\[1em] = \frac{17}{7}\\[1em] = 2\frac{3}{7}\end{aligned}$$

Hence, $\frac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}} = 2\frac{3}{7}$.

Given:

sin θ = $\frac{p}{q}$

i.e. $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{p}{q}$

∴ If length of BC = px unit, length of AC = qx unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ (qx)² = (px)² + AB²

⇒ AB² = q²x² - p²x²

⇒ AB = $\sqrt{q^2\text{x}^2 - p^2\text{x}^2}$

⇒ AB = ($\sqrt{q^2 - p^2}$) x

cos θ = $\frac{Base}{Hypotenuse}$

$= \frac{AB}{AC} = \frac{\sqrt{(q^2 - p^2)}x}{qx} = \frac{\sqrt{q^2 - p^2}}{q}$

Now,

$$\begin{aligned}\text{cos θ} + \text{sin θ} = \frac{\sqrt{q^2 - p^2}}{q} + \frac{p}{q}\\[1em] = \frac{\sqrt{q^2 - p^2} + p}{q}\end{aligned}$$

Hence, cos θ + sin θ = $\frac{\sqrt{q^2 - p^2} + p}{q}$.

Given:

cos $A = \frac{1}{2}$

i.e., $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{2}$

∴ If length of AM = 1x unit, length of AO = 2x unit.

In Δ AMO,

⇒ AO² = AM² + MO² (∵ AC is hypotenuse)

⇒ (2x)² = (1x)² + MO²

⇒ 4x² = 1x² + MO²

⇒ MO² = 4x² - 1x²

⇒ MO² = 3x²

⇒ MO = $\sqrt{3\text{x}^2}$

⇒ MO = $\sqrt{3}$ x

tan $A = \frac{Perpendicular}{Base}$

$= \frac{OM}{MA} = \frac{\sqrt{3} x}{1x} = \frac{\sqrt{3}}{1} = \sqrt{3}$

And,

sin $B = \frac{1}{\sqrt{2}}$

i.e., $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}}$

∴ If length of XY = y unit, length of YB = y $\sqrt{2}$ unit.

In Δ BXY,

⇒ YB² = YX² + BX² (∵ AC is hypotenuse)

⇒ ($\sqrt{2}$y)² = (y)² + BX²

⇒ 2y² = y² + BX²

⇒ BX² = 2y² - y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

tan $B = \frac{Perpendicular}{Base}$

$= \frac{YX}{XB} = \frac{y}{y}= 1$

Now,

$$\begin{aligned}\frac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}\\[1em] = \frac{{\sqrt{3} - 1}}{1 + {\sqrt{3}}}\\[1em] = \frac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \frac{(\sqrt{3} - 3 - 1 + \sqrt{3})}{1^2 - \sqrt{3}^2}\\[1em] = \frac{2 \sqrt{3} - 4}{1 - 3}\\[1em] = \frac{2 (\sqrt{3} - 2)}{-2}\\[1em] = \frac{\cancel{2} (\sqrt{3} - 2)}{-\cancel{2}}\\[1em] = -\sqrt{3} + 2\\[1em] = 2 - \sqrt{3}\end{aligned}$$

Hence, $\frac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}} = 2 - \sqrt{3}$.

Given:

cos $A = \frac{1}{\sqrt2}$

i.e., $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{1}{\sqrt2}$

∴ If length of AM = x unit, length of AO = x $\sqrt{2}$ unit.

In Δ AMO,

⇒ AO² = AM² + MO² (∵ AC is hypotenuse)

⇒ ($\sqrt{2}$x)² = (x)² + MO²

⇒ 2x² = x² + MO²

⇒ MO² = 2x² - x²

⇒ MO² = x²

⇒ MO = $\sqrt{\text{x}^2}$

⇒ MO = x

tan $A = \frac{Perpendicular}{Base}$

$= \frac{OM}{MA} = \frac{x}{x} = 1$

And,

sin $B = \frac{\sqrt3}{2}$

i.e. $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{\sqrt3}{2}$

∴ If length of XY = $\sqrt{3}$ y unit, length of YB = 2y unit.

In Δ BXY,

⇒ YB² = YX² + BX² (∵ AC is hypotenuse)

⇒ (2y)² = ($\sqrt{3}$y)² + BX²

⇒ 4y² = 3y² + BX²

⇒ BX² = 4y² - 3y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

tan $B = \frac{Perpendicular}{Base}$

$= \frac{XY}{BX} = \frac{\sqrt{3}y}{y} = \frac{\sqrt{3}}{1}$

Now,

$$\begin{aligned}\frac{\text{tan B} - \text{tan A}}{1 + \text{tan A } \text{tan B}}\\[1em] = \frac{\frac{\sqrt{3}}{1} - 1}{1 + \frac{\sqrt{3}}{1} \times 1}\\[1em] = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}\\[1em] = \frac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \frac{(\sqrt{3} - 3 - 1 + \sqrt{3})}{1^2 - \sqrt{3}^2}\\[1em] = \frac{2 \sqrt{3} - 4}{1 - 3}\\[1em] = \frac{2 (\sqrt{3} - 2)}{-2}\\[1em] = \frac{\cancel{2} (\sqrt{3} - 2)}{-\cancel{2}}\\[1em] = -\sqrt{3} + 2\\[1em] = 2 - \sqrt{3}\end{aligned}$$

Hence, option 1 is the correct option.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

In Δ ADC,

⇒ DC² = AC² + DA² (∵ DC is hypotenuse)

⇒ DC² = 5² + 12²

⇒ DC² = 25 + 144

⇒ DC² = 169

⇒ DC = $\sqrt{169}$

⇒ DC = 13

cos y = $\frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{DA}{DC}\\[1em] = \frac{12}{13}\end{aligned}$$

Hence, option 3 is the correct option.

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

In Δ AOB,

⇒ AB² = BO² + OA² (∵ AB is hypotenuse)

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

cosec $x = \frac{Hypotenuse}{Perpendicular}$

$$\begin{aligned}= \frac{AB}{OB}\\[1em] = \frac{10}{6}\\[1em] = \frac{5}{3}\\[1em] = 1\frac{2}{3}\end{aligned}$$

Hence, option 1 is the correct option.

Given:

sin A - cosec A = 2

Squaring both sides,

⇒ (sin A - cosec A)² = (2)²

⇒ (sin A)² + (cosec A)² - 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A - 2 x $\text{sin A}$ x $\frac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A - 2 x $\cancel{\text{sin A}}$x $\frac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A - 2 = 4

⇒ sin² A + cosec² A = 4 + 2

⇒ sin² A + cosec² A = 6

Hence, option 4 is the correct option.

Given:

$$\begin{aligned}\text{cot A} = {\sqrt5}\\[1em] \text{cot A} = \frac{\text{Base}}{\text{Perpendicular}} = \frac{\sqrt5}{1}\\[1em]\end{aligned}$$

∴ If length of AB = $\sqrt{5}$ x unit, length of BC = x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = $(\sqrt{5}$ x)² + (x)²

⇒ AC² = 5x² + x²

⇒ AC² = 6x²

⇒ AC = $\sqrt{6\text{x}^2}$

⇒ AC = $\sqrt{6}$ x

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$= \frac{AC}{BC} = \frac{\sqrt{6}\text{x}}{\text{x}} = \frac{\sqrt{6}}{1}$

sec $A = \frac{Hypotenuse}{Base}$

$= \frac{AC}{AB} = \frac{\sqrt{6}\text{x}}{\sqrt{5}\text{x}} = \frac{\sqrt{6}}{\sqrt{5}}$

Now, cosec² A - sec² A

$$\begin{aligned}= \Big(\frac{\sqrt{6}}{1}\Big)^2 - \Big(\frac{\sqrt{6}}{\sqrt{5}}\Big)^2\\[1em] = \frac{6}{1} - \frac{6}{5}\\[1em] = \frac{6 \times 5}{1 \times 5} - \frac{6 \times 1}{5 \times 1}\\[1em] = \frac{30}{5} - \frac{6}{5}\\[1em] = \frac{30 - 6}{5}\\[1em] = \frac{24}{5}\\[1em] = 4\frac{4}{5}\end{aligned}$$

Hence, option 2 is the correct option.

(i) In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ 2² = y² + 1²

⇒ 4 = y² + 1

⇒ y² = 4 - 1

⇒ y² = 3

⇒ y = $\sqrt{3}$

Hence, the value of y = $\sqrt{3}$.

(ii) sin $x° = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{AB}{AC}\\[1em] = \frac{\sqrt{3}}{2}\end{aligned}$$

Hence, sin $x° = \frac{\sqrt{3}}{2}$.

(iii) (sec x° - tan x°)(sec x° + tan x°)

sec $x° = \frac{Hypotenuse}{Base}$

$$\begin{aligned}= \frac{AC}{CB}\\[1em] = \frac{2}{1} = 2\end{aligned}$$

tan $x° = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{AB}{CB}\\[1em] = \frac{\sqrt{3}}{1} = 3\end{aligned}$$

Now, (sec x° - tan x°)(sec x° + tan x°)

$$\begin{aligned}= (2 - \sqrt{3})(2 + \sqrt{3}) \\[1em] = (2)^2 - (\sqrt{3})^2 \\[1em] = 4 - 3 \\[1em] = 1\end{aligned}$$

Hence, (sec x° - tan x°)(sec x° + tan x°) = 1.

In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ BD² = 6² + 8²

⇒ BD² = 36 + 64

⇒ BD² = 100

⇒ BD = $\sqrt{100}$

⇒ BD = 10

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ 17² = AC² + 8²

⇒ 289 = AC² + 64

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) sin $x° = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{DC}{AD}\\[1em] = \frac{8}{17}\end{aligned}$$

Hence, sin $x° = \frac{8}{17}$.

(ii) cos $y° = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{BC}{BD}\\[1em] = \frac{6}{10}\\[1em] = \frac{3}{5}\end{aligned}$$

Hence, cos $y° = \frac{3}{5}$.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan $x° = \frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{DC}{AC}\\[1em] = \frac{8}{15}\end{aligned}$$

sin $y° = \frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{CD}{BD}\\[1em] = \frac{8}{10}\\[1em] = \frac{4}{5}\end{aligned}$$

cos $y° = \frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{BC}{BD}\\[1em] = \frac{6}{10}\\[1em] = \frac{3}{5}\end{aligned}$$

Now, 3 tan x° - 2 sin y° + 4 cos y°

$$\begin{aligned}= 3 \times \frac{8}{15} - 2 \times \frac{4}{5} + 4 \times \frac{3}{5}\\[1em] = \frac{24}{15} - \frac{8}{5} + \frac{12}{5}\\[1em] = \frac{24}{15} + \frac{- 8 + 12}{5}\\[1em] = \frac{24}{15} + \frac{4}{5}\\[1em] = \frac{24}{15} + \frac{4 \times 3}{5 \times 3}\\[1em] = \frac{24}{15} + \frac{12}{15}\\[1em] = \frac{24 + 12}{15}\\[1em] = \frac{36}{15}\\[1em] = \frac{12}{5}\\[1em] = 2\frac{2}{5}\end{aligned}$$

Hence, 3 tan x° - 2 sin y° + 4 cos y° = $2\frac{2}{5}$.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$

⇒ AC = 13

Let ∠CBD be x°.

So, ∠DBA = 90° - x°.

In Δ DAB, according to angle sum property

⇒ ∠ DAB + ∠ADB + ∠DBA = 180°

⇒ ∠ DAB + 90° + (90° - x°) = 180°

⇒ ∠ DAB + 180° - x° = 180°

⇒ ∠ DAB - x° = 0

⇒ ∠ DAB = x°

From figure,

∠ DAB = ∠ CAB = x°

∴ ∠ CBD = ∠ CAB = x°

(i) cos ∠DBC = cos ∠CAB = $\frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{AB}{AC}\\[1em] = \frac{12}{13}\end{aligned}$$

Hence, cos ∠DBC = $\frac{12}{13}$.

(ii) In Δ BCD, according to angle sum property

⇒ ∠ DBC + ∠DCB + ∠CDB = 180°

⇒ ∠ DCB + x° + 90° = 180°

⇒ ∠ DCB = 180° - 90° - x°

⇒ ∠ DCB = 90° - x°

From figure,

∠ DCB = ∠ ACB = 90° - x°

∴ ∠ DBA = ∠ ACB = 90° - x°

cot ∠DBA = cot ∠ACB = $\frac{Base}{Perpendicular}$

$$\begin{aligned}= \frac{BC}{AB}\\[1em] = \frac{5}{12}\end{aligned}$$

Hence, cot ∠DBA = $\frac{5}{12}$.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

Let CD = y and BD = x

In Δ BCD,

⇒ BC² = CD² + BD² (∵ BC is hypotenuse)

⇒ 3² = y² + x²

⇒ 9 = y² + x² ................(1)

In Δ ABD,

⇒ AB² = AD² + BD² (∵ BC is hypotenuse)

⇒ 4² = (5 - y)² + x²

⇒ 16 = 25 + y² - 10y + x² ................(2)

Subtracting (1) from (2), we get

⇒ 16 - 9 = (25 + y² - 10y + x²) - (y² + x²)

⇒ 7 = 25 + y² - 10y + x² - y² - x²

⇒ 7 = 25 - 10y

⇒ 7 - 25 = -10y

⇒ -10y = -18

⇒ y = $\frac{18}{10}$ = 1.8

AD = 5 - 1.8 = 3.2

Using equation (1),

⇒ 9 = (1.8)² + x²

⇒ x² = 9 - 3.24

⇒ x² = 5.76

⇒ x = $\sqrt{5.76}$

⇒ x = 2.4

(i) cos ∠DBC = $\frac{Perpendicular}{Base}$

$$\begin{aligned}= \frac{CD}{BD}\\[1em] = \frac{1.8}{2.4}\\[1em] = \frac{3}{4}\end{aligned}$$

Hence, tan ∠DBC = $\frac{3}{4}$.

(ii) sin ∠DBA = $\frac{Perpendicular}{Hypotenuse}$

$$\begin{aligned}= \frac{AD}{AB}\\[1em] = \frac{3.2}{4}\\[1em] = \frac{4}{5}\end{aligned}$$

Hence, sin ∠DBA = $\frac{4}{5}$.

In isosceles triangle ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\frac{18}{2}$ = 9

cos ∠ABC = $\frac{Base}{Hypotenuse}$

$$\begin{aligned}= \frac{BD}{AB}\\[1em] = \frac{9}{15}\\[1em] = \frac{3}{5}\end{aligned}$$

Hence, cos ∠ABC = $\frac{3}{5}$.

In isosceles Δ ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\frac{8}{2}$ = 4

In Δ ABD,

⇒ AB² = BD² + AD² (∵ AB is hypotenuse)

⇒ 5² = 4² + AD²

⇒ 25 = 16 + AD²

⇒ AD² = 25 - 16

⇒ AD² = 9

⇒ AD = $\sqrt{9}$

⇒ AD = 3

(i) sin B = $\frac{Perpendicular}{Hypotenuse}$

$\frac{AD}{AB} = \frac{3}{5}$

Hence, sin B = $\frac{3}{5}$

(ii) tan C = $\frac{Perpendicular}{Base}$

$\frac{AD}{DC} = \frac{3}{4}$

Hence, tan C = $\frac{3}{4}$

(iii) sin² B + cos² B

sin B = $\frac{Perpendicular}{Hypotenuse}$

$\frac{AD}{AB} = \frac{3}{5}$

cos B = $\frac{Base}{Hypotenuse}$

$\frac{BD}{AB} = \frac{4}{5}$

Now, sin² B + cos² B

$$\begin{aligned}= \Big(\frac{3}{5}\Big)^2 + \Big(\frac{4}{5}\Big)^2\\[1em] = \frac{9}{25} + \frac{16}{25}\\[1em] = \frac{9 + 16}{25}\\[1em] = \frac{25}{25}\\[1em] = 1\end{aligned}$$

Hence, sin² B + cos² B = 1.

(iv) tan C - cot B

tan C = $\frac{Perpendicular}{Base}$

$= \frac{AD}{DC} = \frac{3}{4}$

cot B = $\frac{Base}{Perpendicular}$

$= \frac{BD}{AD} = \frac{4}{3}$

Now, tan C - cot B

$$\begin{aligned}= \frac{3}{4} - \frac{4}{3}\\[1em] = \frac{3 \times 3}{4 \times 3}- \frac{4 \times 4}{3 \times 4}\\[1em] = \frac{9}{12} - \frac{16}{12}\\[1em] = \frac{9 - 16}{12}\\[1em] = \frac{-7}{12}\end{aligned}$$

Hence, tan C - cot B = $\frac{-7}{12}$.

Given:

$$\begin{aligned}\text{tan x°} = \frac{3}{4}\\[1em] \text{tan x°} = \frac{Perpendicular}{Base} = \frac{3}{4} \\[1em] \Rightarrow \frac{BC}{AB} = \frac{3}{4}\end{aligned}$$

∴ If length of AB = 4x unit, length of BC = 3x unit.

BC = 15 cm (∵ Given)

∴ 3x = 15

⇒ x = $\frac{15}{3}$ = 5 cm

∴ AB = 4x = 4 x 5 = 20 cm

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AB is hypotenuse)

⇒ AC² = (15)² + (20)²

⇒ AC² = 225 + 400

⇒ AC² = 625

⇒ AC = $\sqrt{625}$

⇒ AC = 25 cm

Hence, AB = 20 cm and AC = 25 cm.

(i) In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ 13² = 12² + CD²

⇒ 169 = 144 + CD²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$

⇒ CD = 5

sin Φ = $\frac{Perpendicular}{Hypotenuse}$

$= \frac{CD}{BD} = \frac{5}{13}$

Draw a line parallel to BC from point D such that it meets AB at point E. This line DE will be ⊥ to AB.

From figure,

DE = BC = 12

In Δ BED,

⇒ BD² = BE² + DE² (∵ BD is hypotenuse)

⇒ 13² = BE² + 12²

⇒ 169 = BE² + 144

⇒ BE² = 169 - 144

⇒ BE² = 25

⇒ BE = $\sqrt{25}$

⇒ BE = 5

And, AE = AB - BE = 14 - 5 = 9

tan θ = $\frac{Base}{Perpendicular}$

$= \frac{DE}{AE} = \frac{12}{9} = \frac{4}{3}$

Hence, sin Φ = $\frac{5}{13}$ and tan θ = $\frac{4}{3}$

(ii) sin θ = $\frac{Perpendicular}{Hypotenuse}$

sin θ = $\frac{DE}{AD} = \frac{12}{AD}$

AD = $\frac{12}{\text{sin θ}}$ = 12 cosec θ

cos θ = $\frac{Base}{Hypotenuse}$

cos θ = $\frac{AE}{AD} = \frac{9}{AD}$

AD = $\frac{9}{\text{cos θ}}$ = 9 sec θ

Hence, AD = 12 cosec θ or 9 sec θ.

(i) Given:

$$\begin{aligned}\text{sin B} = \frac{4}{5}\\[1em] \text{sin B} = \frac{Perpendicular}{Hypotenuse} = \frac{4}{5}\end{aligned}$$

∴ If length of AC = 4x cm, length of AB = 5x cm.

In Δ ABC,

⇒ AB²= BC² + AC² (∵ AB is hypotenuse)

⇒ (5x)² = BC² + (4x)²

⇒ 25x² = BC² + 16x²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = $\sqrt{9\text{x}^2}$

⇒ BC = 3x

It is given that BC = 15 cm

3x = 15

x = $\frac{15}{3}$

x = 5 cm

AB = 5x = 5 x 5 cm = 25 cm

AC = 4x = 4 x 5 cm = 20 cm

Hence, AB = 25 cm and AC = 20 cm.

(ii) tan ∠ADC = 1

$\text{tan ∠ADC} = \frac{Perpendicular}{Base} = 1$

∴ If length of AC = x unit, length of CD = x unit.

From (i), we know AC = 20 cm

∴ x = 20 cm

So, AC = CD = 20 cm

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ AD² = 20² + 20²

⇒ AD² = 400 + 400

⇒ AD² = 800

⇒ AD = $\sqrt{800}$

⇒ AD = $20\sqrt{2}$

Hence, CD = 20 cm and AD = 20 $\sqrt{2}$ cm.

sin A + cosec A = 2

Squaring both sides,

(sin A + cosec A)² = 2²

⇒ sin² A + cosec² A + 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A + 2 x sin A x $\frac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A + 2 x ${\cancel{\text{sin A}}}$x $\frac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A + 2 = 4

⇒ sin² A + cosec² A = 4 - 2

⇒ sin² A + cosec² A = 2

Hence, sin² A + cosec² A = 2.

tan A + cot A = 5

Squaring both sides,

(tan A + cot A)² = 5²

⇒ tan² A + cot² A + 2 x tan A x cot A = 25

⇒ tan² A + cot² A + 2 x tan A x $\frac{1}{\text{tan A}}$ = 25

⇒ tan² A + cot² A + 2 x ${\cancel{\text{tan A}}}$x $\frac{1}{\cancel{\text{tan A}}}$ = 25

⇒ tan² A + cot² A + 2 = 25

⇒ tan² A + cot² A = 25 - 2

⇒ tan² A + cot² A = 23

Hence, tan² A + cot² A = 23.

Given:

4 sin θ = 3 cos θ

$$\begin{aligned}⇒ \frac{\text{sin θ}}{\text{cos θ}} = \frac{3}{4}\\[1em] ⇒ \text{tan θ} = \frac{3}{4}\\[1em] ⇒ \text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] ⇒ \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4}\\[1em]\end{aligned}$$

∴ If length of BC = 3x unit, length of AB = 4x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25 \text{x}^2}$

⇒ AC = 5x

(i) sin θ = $\frac{Perpendicular}{Hypotenuse}$

$\frac{CB}{AC} = \frac{3x}{5x} = \frac{3}{5}$

Hence, sin θ = $\frac{3}{5}$.

(ii) cos θ = $\frac{Base}{Hypotenuse}$

$\frac{AB}{AC} = \frac{4x}{5x} = \frac{4}{5}$

Hence, cos θ = $\frac{4}{5}$.

(iii) cot² θ - cosec² θ + 2

cot θ = $\frac{Base}{Perpendicular}$

$\frac{AB}{BC} = \frac{4x}{3x} = \frac{4}{3}$

cosec θ = $\frac{Hypotenuse}{Perpendicular}$

$\frac{AC}{CB} = \frac{5x}{3x} = \frac{5}{3}$

Now, cot² θ - cosec² θ

$$\begin{aligned}= \Big(\frac{4}{3}\Big)^2 - \Big(\frac{5}{3}\Big)^2 \\[1em] = \frac{16}{9} - \frac{25}{9} \\[1em] = \frac{16 - 25}{9} \\[1em] = \frac{- 9}{9} \\[1em] = -1\end{aligned}$$

Hence, cot² θ - cosec² θ + 2 = -1.

(iv) 4 cos² θ - 3 sin² θ + 2

cos θ = $\frac{Base}{Hypotenuse}$

$\frac{AB}{AC} = \frac{4x}{5x} = \frac{4}{5}$

sin θ = $\frac{Perpendicular}{Hypotenuse}$

$\frac{CB}{AC} = \frac{3x}{5x} = \frac{3}{5}$

Now, 4 cos² θ - 3 sin² θ + 2

$$\begin{aligned}= 4 \times \Big(\frac{4}{5}\Big)^2 - 3 \times \Big(\frac{3}{5}\Big)^2 + 2\\[1em] = 4 \times \frac{16}{25} - 3 \times \frac{9}{25} + 2\\[1em] = \frac{64}{25} - \frac{27}{25} + 2\\[1em] = \frac{64 - 27}{25} + 2\\[1em] = \frac{37}{25} + 2\\[1em] = \frac{37}{25} + \frac{2 \times 25}{25}\\[1em] = \frac{37}{25} + \frac{50}{25}\\[1em] = \frac{37 + 50}{25}\\[1em] = \frac{87}{25}\\[1em] = 3\frac{12}{25}\end{aligned}$$

Hence, 4 cos² θ - 3 sin² θ + 2 =$3\frac{12}{25}$.

Given:

17 cos θ = 15

cos θ = $\frac{15}{17}$

$$\begin{aligned}\Rightarrow \text{cos θ} = \frac{Base}{Hypotenuse} = \frac{15}{17}\\[1em]\end{aligned}$$

∴ If length of AC = 17x unit, length of AB = 15x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ (17x)² = BC² + (15x)²

⇒ 289x² = BC² + 225x²

⇒ BC² = 289x² - 225x²

⇒ BC = $\sqrt{64 \text{x}^2}$

⇒ BC = 8x

tan θ = $\frac{Perpendicular}{Base}$

$= \frac{CB}{AB} = \frac{8x}{15x} = \frac{8}{15}$

sec θ = $\frac{Hypotenuse}{Base}$

$= \frac{AC}{AB} = \frac{17x}{15x} = \frac{17}{15}$

Now, tan θ + 2 sec θ

$$\begin{aligned}= \frac{8}{15} + 2 \times \frac{17}{15}\\[1em] = \frac{8}{15} + \frac{34}{15}\\[1em] = \frac{8 + 34}{15}\\[1em] = \frac{42}{15}\\[1em] = \frac{14}{5}\\[1em] = 2\frac{4}{5}\end{aligned}$$

Hence, tan θ + 2 sec θ = $2\frac{4}{5}$.

Given:

5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

⇒ $\frac{\text{sin A}}{\text{cos A}} = \frac{5}{12}$

⇒ $\text{tan A} = \frac{5}{12}$

$$\begin{aligned}⇒ \text{tan A} = \frac{Perpendicular}{Base} = \frac{5}{12} \\[1em]\end{aligned}$$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169 \text{x}^2}$

⇒ AC = 13x

sin A = $\frac{Perpendicular}{Hypotenuse}$

$\frac{CB}{AC} = \frac{5x}{13x} = \frac{5}{13}$

cos A = $\frac{Base}{Hypotenuse}$

$\frac{AB}{AC} = \frac{12x}{13x} = \frac{12}{13}$

Now,

$$\begin{aligned}\frac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}\\[1em] = \frac{\frac{5}{13} + \frac{12}{13}}{2 \times \frac{12}{13} - \frac{5}{13}}\\[1em] = \frac{\frac{5 + 12}{13}}{\frac{24}{13} - \frac{5}{13}}\\[1em] = \frac{\frac{17}{13}}{\frac{24 - 5}{13}}\\[1em] = \frac{\frac{17}{13}}{\frac{19}{13}}\\[1em] = \frac{\frac{17}{\cancel{13}}}{\frac{19}{\cancel{13}}}\\[1em] = \frac{17}{19}\end{aligned}$$

Hence, $\frac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}} = \frac{17}{19}$.

$$\begin{aligned}\frac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \Big(\frac{1}{\sqrt{3}}\Big)^2}\\[1em] = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt{3}}}{\frac{3 + 1}{3}}\\[1em] = \frac{2 \times 3}{4 \times {\sqrt{3}}}\\[1em] = \frac{6}{4{\sqrt{3}}}\\[1em] = \frac{3}{2{\sqrt{3}}}\\[1em] = \frac{{\sqrt{3}}}{2}\\[1em] = \text{sin} 60°\end{aligned}$$

Hence, option 1 is the correct option.

tan 3A - ${\sqrt3}$ = 0

tan 3A = ${\sqrt3}$

tan 3A = tan 60°

So, 3A = 60°

A = $\frac{60°}{3}$

A = 20°

Hence, option 2 is the correct option.

cot A = tan A

⇒ $\frac{1}{\text{tan A}}$ = tan A

⇒ tan² A = 1

⇒ tan A = 1

⇒ tan A = tan 45°

Hence, A = 45°

Hence, option 3 is the correct option.

$$\begin{aligned}\text{cos}^2 60° - 2 \text{sin}^3 30° + 3 \text{cot}^4 45° = \Big(\frac{1}{2}\Big)^2 - 2 \times \Big(\frac{1}{2}\Big)^3 + 3 \times (1)^2\\[1em] = \frac{1}{4} - 2 \times \frac{1}{8} + 3\\[1em] = \frac{1}{4} - \frac{1}{4} + 3\\[1em] = 3\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}\frac{\text{cos 60° - cos 0°} + \text{2 sin 90°}}{\text{ cot 60°}\times \text{ cot 30°}} = \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt3}\times \sqrt3}\\[1em] = \frac{\frac{1}{2} - 1 + 2}{1}\\[1em] = \frac{1}{2} + 1\\[1em] = \frac{1 + 2}{2}\\[1em] = \frac{3}{2}\\[1em] = 1\frac{1}{2}\end{aligned}$$

Hence, option 1 is the correct option.

$$\begin{aligned}\text{sin 30° cos 30°} = \frac{1}{2} \times \frac{\sqrt3}{2}\\[1em] = \frac{\sqrt3}{4}\\[1em]\end{aligned}$$

Hence, sin 30° cos 30° = $\frac{\sqrt3}{4}$.

$$\begin{aligned}\text{tan 30° tan 60°} = \frac{1}{\sqrt3} \times \sqrt3\\[1em] = \frac{1}{\cancel{\sqrt3}} \times \cancel{\sqrt3}\\[1em] = 1\end{aligned}$$

Hence, tan 30° tan 60° = 1.

$$\begin{aligned}\text{cos}^2 60° + \text{sin}^2 30° = \Big(\frac{1}{2}\Big)^2 + \Big(\frac{1}{2}\Big)^2\\[1em] = \frac{1}{4} + \frac{1}{4}\\[1em] = \frac{1 + 1}{4}\\[1em] = \frac{2}{4}\\[1em] = \frac{1}{2}\end{aligned}$$

Hence, cos² 60° + sin² 30° = $\frac{1}{2}$.

$$\begin{aligned}\text{cosec}^2 60° - \text{tan}^2 30° = \Big(\frac{2}{\sqrt3}\Big)^2 - \Big(\frac{1}{\sqrt3}\Big)^2\\[1em] = \frac{4}{3} - \frac{1}{3}\\[1em] = \frac{4 - 1}{3}\\[1em] = \frac{3}{3}\\[1em] = 1\end{aligned}$$

Hence, cosec² 60° - tan² 30° = 1.

$$\begin{aligned}\text{sin}^2 30° + \text{cos}^2 30° + \text{cot}^2 45° = \Big(\frac{1}{2}\Big)^2 + \Big(\frac{\sqrt3}{2}\Big)^2 + (1)^2\\[1em] = \frac{1}{4} + \frac{3}{4} + 1\\[1em] = \frac{1 + 3}{4} + 1\\[1em] = \frac{4}{4} + 1\\[1em] = 1 + 1\\[1em] = 2\end{aligned}$$

Hence, sin² 30° + cos² 30° + cot² 45° = 2.

$$\begin{aligned}\text{cos}^2 60° + \text{sec}^2 30° + \text{tan}^2 45° = \Big(\frac{1}{2}\Big)^2 + \Big(\frac{2}{\sqrt3}\Big)^2 + (1)^2\\[1em] = \frac{1}{4} + \frac{4}{3} + 1\\[1em] = \frac{1 \times 3}{4 \times 3} + \frac{4 \times 4}{3 \times 4} + \frac{1 \times 12}{12}\\[1em] = \frac{3}{12} + \frac{16}{12} + \frac{12}{12}\\[1em] = \frac{3 + 16 + 12}{12}\\[1em] = \frac{31}{12}\\[1em] = 2\frac{7}{12}\end{aligned}$$

Hence, cos² 60° + sec² 30° + tan² 45° = $2\frac{7}{12}$.

tan² 30° + tan² 45° + tan² 60°

$$\begin{aligned}= \Big(\frac{1}{\sqrt3}\Big)^2 + (1)^2 + (\sqrt3)^2\\[1em] = \frac{1}{3} + 1 + 3\\[1em] = \frac{1}{3} + \frac{1 \times 3}{3} + \frac{3 \times 3}{3}\\[1em] = \frac{1}{3} + \frac{3}{3} + \frac{9}{3}\\[1em] = \frac{1 + 3 + 9}{3}\\[1em] = \frac{13}{3}\\[1em] = 4\frac{1}{3}\end{aligned}$$

Hence, tan² 30° + tan² 45° + tan² 60° = $4\frac{1}{3}$.

$\frac{\text{tan 45°}}{\text{cosec 30°}} + \frac{\text{sec 60°}}{\text{cot 45°}} - \frac{\text{5 sin 90°}}{\text{2 cos 0°}}$

$$\begin{aligned}= \frac{1}{2} + \frac{2}{1} - \frac{5 \times 1}{2 \times 1}\\[1em] = \frac{1}{2} + \frac{2 \times 2}{1 \times 2} - \frac{5}{2}\\[1em] = \frac{1}{2} + \frac{4}{2} - \frac{5}{2}\\[1em] = \frac{1 + 4 - 5}{2}\\[1em] = 0\end{aligned}$$

Hence, $\frac{\text{tan 45°}}{\text{cosec 30°}} + \frac{\text{sec 60°}}{\text{cot 45°}} - \frac{\text{5 sin 90°}}{\text{2 cos 0°}} = 0$

3 sin² 30° + 2 tan² 60° - 5 cos² 45°

$$\begin{aligned}= 3 \times \Big(\frac{1}{2}\Big)^2 + 2 \times (\sqrt3)^2 - 5 \times \Big(\frac{1}{\sqrt2}\Big)^2\\[1em] = 3 \times \frac{1}{4} + 2 \times 3 - 5 \times \frac{1}{2}\\[1em] = \frac{3}{4} + 6 - \frac{5}{2}\\[1em] = \frac{3}{4} + \frac{6 \times 4}{4} - \frac{5 \times 2}{2 \times 2}\\[1em] = \frac{3}{4} + \frac{24}{4} - \frac{10}{4}\\[1em] = \frac{3 + 24 - 10}{4}\\[1em] = \frac{17}{4}\\[1em] = 4\frac{1}{4}\end{aligned}$$

Hence, 3 sin² 30° + 2 tan² 60° - 5 cos² 45° = $4\frac{1}{4}$.

sin 60° cos 30° + cos 60°. sin 30° = 1

L.H.S = sin 60° cos 30° + cos 60°. sin 30°

$$\begin{aligned}= \frac{\sqrt3}{2} \times \frac{\sqrt3}{2} + \frac{1}{2} \times \frac{1}{2} \\[1em] = \frac{3}{4} + \frac{1}{4}\\[1em] = \frac{3 + 1}{4}\\[1em] = \frac{4}{4}\\[1em] = 1\end{aligned}$$

R.H.S = 1

∴ L.H.S = R.H.S

Hence proved, sin 60° cos 30° + cos 60°. sin 30° = 1.

cos 30°. cos 60° - sin 30°. sin 60° = 0

L.H.S. = cos 30°. cos 60° - sin 30°. sin 60°

$$\begin{aligned}= \frac{\sqrt3}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt3}{2} \\[1em] = \frac{\sqrt3}{4} - \frac{\sqrt3}{4}\\[1em] = 0\end{aligned}$$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence proved, cos 30°. cos 60° - sin 30°. sin 60° = 0.

cosec² 45° - cot² 45° = 1

L.H.S. = cosec² 45° - cot² 45°

$$\begin{aligned}= (\sqrt2)^2 - (1)^2 \\[1em] = 2 - 1\\[1em] = 1\end{aligned}$$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence proved, cosec² 45° - cot² 45° = 1.

cos² 30° - sin² 30° = cos 60°

L.H.S. = cos² 30° - sin² 30°

$$\begin{aligned}= \Big(\frac{\sqrt3}{4}\Big)^2 - \Big(\frac{1}{4}\Big)^2 \\[1em] = \frac{3}{4} - \frac{1}{4}\\[1em] = \frac{3 - 1}{4}\\[1em] = \frac{2}{4}\\[1em] = \frac{1}{2}\end{aligned}$$

R.H.S.. = cos 60° = $\frac{1}{2}$

∴ L.H.S. = R.H.S.

Hence proved, cos² 30° - sin² 30° = cos 60°.

$\Big(\frac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \frac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

$$\begin{aligned}\text{L.H.S.} = \Big(\frac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2\\[1em] = \Big(\frac{\sqrt3 + 1}{\sqrt3 - 1}\Big)^2\\[1em] = \Big(\frac{(\sqrt3 + 1) \times (\sqrt3 + 1)}{(\sqrt3 - 1) \times (\sqrt3 + 1)}\Big)^2\\[1em] = \Big(\frac{(\sqrt3 + 1)^2}{(\sqrt3)^2 - (1)^2}\Big)^2\\[1em] = \Big(\frac{3 + 1 + 2 \times 1 \times \sqrt3}{3 - 1}\Big)^2\\[1em] = \Big(\frac{4 + 2\sqrt3}{2}\Big)^2\\[1em] = (2 + \sqrt3)^2\\[1em] = 4 + 3 + 2 \times 2 \times \sqrt3\\[1em] = 7 + 4\sqrt3\end{aligned}$$

$$\begin{aligned}\text{R.H.S.} = \frac{\text{1 + cos 30°}}{\text{1 - cos 30°}}\\[1em] = \frac{1 + \frac{\sqrt3}{2}}{1 - \frac{\sqrt3}{2}}\\[1em] = \frac{\frac{2 + \sqrt3}{2}}{\frac{2 - \sqrt3}{2}}\\[1em] = \frac{\frac{2 + \sqrt3}{\cancel{2}}}{\frac{2 - \sqrt3}{\cancel{2}}}\\[1em] = \frac{2 + \sqrt3}{2 - \sqrt3}\\[1em] = \frac{(2 + \sqrt3) \times (2 + \sqrt3)}{(2 - \sqrt3) \times (2 + \sqrt3)}\\[1em] = \frac{(2 + \sqrt3)^2}{(2)^2 - (\sqrt3)^2}\\[1em] = \frac{4 + 3 + 2 \times 2 \times \sqrt3}{4 - 3}\\[1em] = 7 + 4\sqrt3\\[1em]\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, $\Big(\frac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \frac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

L.H.S. = 3 cosec² 60° - 2 cot² 30° + sec² 45°

$$\begin{aligned}= 3 \times \Big(\frac{2}{\sqrt3}\Big)^2 - 2 \times ({\sqrt3})^2 + ({\sqrt2})^2\\[1em] = 3 \times \Big(\frac{4}{3}\Big) - 2 \times 3 + 2\\[1em] = 4 - 6 + 2\\[1em] = 0\end{aligned}$$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence, 3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

$\text{sin }(2 \times 30°) = \frac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = sin (2 x 30°) = sin 60° = $\frac{\sqrt3}{2}$

R.H.S.

$$\begin{aligned}= \frac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \frac{2 \times \frac{1}{\sqrt3}}{1 + \Big(\frac{1}{\sqrt3}\Big)^2}\\[1em] = \frac{\frac{2}{\sqrt3}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{3 + 1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{4}{3}}\\[1em] = \frac{2 \times 3}{4 \times \sqrt3}\\[1em] = \frac{\sqrt3}{2}\\[1em]\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, $\text{sin }(2 \times 30°) = \frac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

$\text{cos }(2 \times 30°) = \frac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = cos (2 x 30°) = cos 60° = $\frac{1}{2}$

R.H.S.

$$\begin{aligned}= \frac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \frac{1 - \Big(\frac{1}{\sqrt3}\Big)^2}{1 + \Big(\frac{1}{\sqrt3}\Big)^2}\\[1em] = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{3}{3} - \frac{1}{3}}{\frac{3}{3} + \frac{1}{3}}\\[1em] = \frac{\frac{3 - 1}{3}}{\frac{3 + 1}{3}}\\[1em] = \frac{\frac{2}{3}}{\frac{4}{3}}\\[1em] = \frac{\frac{2}{\cancel{3}}}{\frac{4}{\cancel{3}}}\\[1em] = \frac{2}{4}\\[1em] = \frac{1}{2}\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, $\text{cos }(2 \times 30°) = \frac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

$\text{tan} (2 \times 30°) = \frac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

L.H.S. = tan 2 x 30° = tan 60° = $\sqrt3$

R.H.S.

$$\begin{aligned}= \frac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}\\[1em] = \frac{2 \times \frac{1}{\sqrt3}}{1 - \Big(\frac{1}{\sqrt3}\Big)^2}\\[1em] = \frac{\frac{2}{\sqrt3}}{1 - \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{3}{3} - \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{3 - 1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{2}{3}}\\[1em] = \frac{2 \times 3}{2 \times \sqrt3}\\[1em] = \sqrt3\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, $\text{tan} (2 \times 30°) = \frac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

In Δ ABC,

⇒ AC² = BC² + AB² (∴ AC is hypotenuse)

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = $\sqrt{2}$ x

As Δ ABC is isosceles right angled triangle, ∠BAC = ∠BCA = $\frac{90°}{2}$ = 45°

(i) sin 45°

sin 45° = sin A = sin C

sin A = $\frac{Perpendicular}{Hypotenuse}$

= $\frac{BC}{AC} = \frac{x}{\sqrt2x} = \frac{1}{\sqrt2}$

Hence, sin 45° = $\frac{1}{\sqrt2}$.

(ii) cos 45°

cos 45° = cos A = cos C

cos A = $\frac{Base}{Hypotenuse}$

= $\frac{AB}{AC} = \frac{x}{\sqrt2x} = \frac{1}{\sqrt2}$

Hence, cos 45° = $\frac{1}{\sqrt2}$.

(iii) tan 45°

tan 45° = tan A = tan C

tan A = $\frac{Perpendicular}{Base}$

= $\frac{BC}{AB} = \frac{x}{x}$ = 1

Hence, tan 45° = 1.

sin 60° = 2 sin 30° cos 30°.

L.H.S. = sin 60° = $\frac{\sqrt3}{2}$

R.H.S.= 2 sin 30° cos 30°

$$\begin{aligned}= 2 \times \frac{1}{2} \times \frac{\sqrt3}{2}\\[1em] = \cancel{2} \times \frac{1}{\cancel{2}} \times \frac{\sqrt3}{2}\\[1em] = \frac{\sqrt3}{2}\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence proved, sin 60° = 2 sin 30° cos 30°.

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2

L.H.S. = 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

$$\begin{aligned}= 4 \times \Big(\Big(\frac{1}{2}\Big)^4 + \Big(\frac{1}{2}\Big)^4\Big) - 3 \times \Big(\Big(\frac{1}{\sqrt2}\Big)^2 - (1)^2\Big)\\[1em] = 4 \times \Big(\frac{1}{16} + \frac{1}{16}\Big) - 3 \times \Big(\frac{1}{2} - 1\Big)\\[1em] = 4 \times \Big(\frac{1 + 1}{16}\Big) - 3 \times \Big(\frac{1}{2} - \frac{2}{2}\Big)\\[1em] = 4 \times \Big(\frac{2}{16}\Big) - 3 \times \Big(\frac{1 - 2}{2}\Big)\\[1em] = 4 \times \Big(\frac{1}{8}\Big) - 3 \times \Big(\frac{-1}{2}\Big)\\[1em] = \frac{1}{2} + \frac{3}{2}\\[1em] = \frac{1 + 3}{2}\\[1em] = \frac{ 4}{2}\\[1em] = 2\end{aligned}$$

R.H.S. = 2

∴ L.H.S. = R.H.S.

Hence, 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2.

sin x = cos x

As we know that sin² x + cos² x = 1

⇒ sin² x + sin² x = 1

⇒ 2sin² x = 1

⇒ sin² x = $\frac{1}{2}$

⇒ sin x = $\frac{1}{\sqrt2}$

⇒ sin x = sin 45°

⇒ x = 45°

Hence, the value of x = 45°.

sec A = cosec A

$\frac{1}{\text{cos A}} = \frac{1}{\text{sin A}}$

sin A = cos A

As we know that sin² A + cos² A = 1

⇒ sin² A + sin² A = 1

⇒ 2sin² A = 1

⇒ sin² A = $\frac{1}{2}$

⇒ sin A = $\frac{1}{\sqrt2}$

⇒ sin A = sin 45°

⇒ A = 45°

Hence, the value of A = 45°.

tan θ = cot θ

$\frac{\text{sin θ}}{\text{cos θ}} = \frac{\text{cos θ}}{\text{sin θ}}$

$\text{sin}^2 \text{θ} = \text{cos}^2 \text{θ}$

As we know that sin² θ + cos² θ = 1

⇒ sin² θ + sin² θ = 1

⇒ 2sin² θ = 1

⇒ sin² θ = $\frac{1}{2}$

⇒ sin θ = $\frac{1}{\sqrt2}$

⇒ sin θ = sin 45°

⇒ θ = 45°

Hence, the value of θ = 45°.

sin x = cos y

cos y = sin (90° - y)

So, sin x = sin (90° - y)

⇒ x = 90° - y

⇒ x + y = 90°

Hence, the relation between x and y is x + y = 90°.

False

Reason:

sin x = cos y

It is only possible only when x = y = 45°.

sin 45° = cos 45° = 1

True

Reason:

sec θ. cot θ = cosec θ

L.H.S. = sec θ. cot θ

$$\begin{aligned}= \frac{1}{\text{cos θ}} \times \frac{\text{cos θ}}{\text{sin θ}}\\[1em] = \frac{1}{\cancel{\text{cos θ}}} \times \frac{\cancel{\text{cos θ}}}{\text{sin θ}}\\[1em] = \frac{1}{\text{sin θ}}\\[1em] = \text{cosec θ}\end{aligned}$$

R.H.S. = cosec θ

∴ L.H.S. = R.H.S.

1

Reason:

sin² θ + cos² θ = 1

Let take θ = 30°

sin² 30° = $\Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$

cos² 30° = $\Big(\frac{\sqrt3}{2}\Big)^2 = \frac{3}{4}$

Now, sin² 30° + cos² 30°

$$\begin{aligned}= \frac{1}{4} + \frac{3}{4}\\[1em] = \frac{1 + 3}{4}\\[1em] = \frac{4}{4}\\[1em] = 1\end{aligned}$$

sin θ increases as θ increases

Reason — As we know,

sin 0° = 0

sin 30° = $\frac{1}{2}$ = 0.5

sin 45° = $\frac{1}{\sqrt2}$ = 0.70

sin 60° = $\frac{\sqrt3}{2}$ = 0.87

sin 90° = 1

cos θ decreases as θ increases

Reason — As we know,

cos 0° = 1

cos 30° = $\frac{\sqrt3}{2}$ = 0.87

cos 45° = $\frac{1}{\sqrt2}$ = 0.70

cos 60° = $\frac{1}{2}$ = 0.5

cos 90° = 0

tan θ decreases as θ decreases.

Reason — As we know,

tan 90° = not defined

tan 60° = $\sqrt3$ = 1.73

tan 45° = 1

tan 30° = $\frac{1}{\sqrt3}$ = 0.58

tan 0° = 0

$$\begin{aligned}\frac{\text{sin 2 A}}{\text{1 - cos 2 A}}\\[1em] = \frac{\text{sin (2 x 30°)}}{\text{1 - cos (2 x 30°)}}\\[1em] = \frac{\text{sin 60°}}{\text{1 - cos 60°}}\\[1em] = \frac{\frac{\sqrt3}{2}}{1 - \frac{1}{2}}\\[1em] = \frac{\frac{\sqrt3}{2}}{\frac{2}{2} - \frac{1}{2}}\\[1em] = \frac{\frac{\sqrt3}{2}}{\frac{2 - 1}{2}}\\[1em] = \frac{\frac{\sqrt3}{2}}{\frac{1}{2}}\\[1em] = \frac{\frac{\sqrt3}{\cancel{2}}}{\frac{1}{\cancel{2}}}\\[1em] = \frac{\sqrt3}{1}\\[1em] = \text {cot A}\end{aligned}$$

Hence, option 1 is the correct option.

sin A cos B + cos A sin B = sin 60°. cos 30° + cos 60°. sin 30°

$$\begin{aligned}= \frac{\sqrt3}{2} \times \frac{\sqrt3}{2} + \frac{1}{2} \times \frac{1}{2}\\[1em] = \frac{3}{4} + \frac{1}{4}\\[1em] = \frac{3 + 1}{4} \\[1em] = \frac{4}{4} \\[1em] = 1\end{aligned}$$

Hence, option 2 is the correct option.

3 sin A - 4 sin³ A = 3 sin 30° - 4 sin³ 30°

$$\begin{aligned}= 3 \times \frac{1}{2} - 4 \times \Big(\frac{1}{2}\Big)^2\\[1em] = \frac{3}{2} - 4 \times \frac{1}{8}\\[1em] = \frac{3}{2} - \frac{1}{2}\\[1em] = \frac{3 - 1}{2} \\[1em] = \frac{2}{2} \\[1em] = 1\\[1em] = \text{sin 90°}\\[1em] = \text{sin 3 x 30°}\\[1em] = \text{sin 3A}\\[1em]\end{aligned}$$

Hence, option 3 is the correct option.

cos⁴ A - sin⁴ A = cos⁴ 30° - sin⁴ 30°

$$\begin{aligned}= \Big(\frac{\sqrt3}{2}\Big)^4 - \Big(\frac{1}{2}\Big)^4\\[1em] = \frac{9}{16} - \frac{1}{16}\\[1em] = \frac{9 - 1}{16}\\[1em] = \frac{8}{16}\\[1em] = \frac{1}{2}\\[1em] = \text {cos 60°}\end{aligned}$$

Hence, option 4 is the correct option.

2 sin A cos A = 2 sin 45°. cos 45°

$$\begin{aligned}= 2 \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt2}\\[1em] = 2 \times \frac{1}{2}\\[1em] = \cancel{2} \times \frac{1}{\cancel{2}}\\[1em] = 1\end{aligned}$$

Hence, option 1 is the correct option.

sin (A + B) = sin A cos B + cos A sin B

L.H.S. = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

$$\begin{aligned}= \frac{\sqrt3}{2} \times \frac{\sqrt3}{2} + \frac{1}{2} \times \frac{1}{2}\\[1em] = \frac{3}{4} + \frac{1}{4}\\[1em] = \frac{3 + 1}{4}\\[1em] = \frac{4}{4}\\[1em] = 1\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, sin (A + B) = sin A cos B + cos A sin B.

cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos (A + B) = cos (60° + 30°)

= cos 90° = 0

R.H.S. = cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

$$\begin{aligned}= \frac{1}{2} \times \frac{\sqrt3}{2} - \frac{\sqrt3}{2} \times \frac{1}{2}\\[1em] = \frac{\sqrt3}{4} - \frac{\sqrt3}{4}\\[1em] = 0\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, cos (A + B) = cos A cos B - sin A sin B.

cos (A - B) = cos A cos B + sin A sin B

L.H.S. = cos (A - B) = cos (60° - 30°)

= cos 30° = $\frac{\sqrt3}{2}$

R.H.S. = cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$$\begin{aligned}= \frac{1}{2} \times \frac{\sqrt3}{2} + \frac{\sqrt3}{2} \times \frac{1}{2}\\[1em] = \frac{\sqrt3}{4} + \frac{\sqrt3}{4}\\[1em] = 2 \times \frac{\sqrt3}{4}\\[1em] = \frac{\sqrt3}{2}\\[1em]\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, cos (A - B) = cos A cos B + sin A sin B.

$\text{tan (A - B)} = \frac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$

L.H.S. = tan (A - B) = tan (60° - 30°)

= tan 30° = $\frac{1}{\sqrt3}$

$$\begin{aligned}\text{R.H.S.} = \frac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}\\[1em] = \frac{\text{tan 60° - tan 30°}}{\text{1 + tan} \text{ 60° } . \text{ tan 30°}}\\[1em] = \frac{\sqrt3 - \frac{1}{\sqrt3}}{1 + \sqrt3 \times \frac{1}{\sqrt3}}\\[1em] = \frac{\frac{\sqrt3 \times \sqrt3}{\sqrt3} - \frac{1}{\sqrt3}}{1 + \cancel{\sqrt3} \times \frac{1}{\cancel{\sqrt3}}}\\[1em] = \frac{\frac{3}{\sqrt3} - \frac{1}{\sqrt3}}{1 + 1}\\[1em] = \frac{\frac{3 - 1}{\sqrt3}}{1 + 1}\\[1em] = \frac{\frac{2}{\sqrt3}}{2}\\[1em] = \frac{\frac{\cancel{2}}{\sqrt3}}{\cancel{2}}\\[1em] = \frac{1}{\sqrt3}\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, $\text{tan (A - B)} = \frac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$.

$\text{sin 2 A} = \text{2 sin A cos A} = \frac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{sin 2 A} = \text{sin (2 x 30°)} = \text{sin 60°} = \frac{\sqrt3}{2}$

$$\begin{aligned}\text{2nd term} = \text{2 sin A cos A}\\[1em] = 2 \times \text{sin 30°} \times \text{cos 30°}\\[1em] = 2 \times \frac{1}{2} \times \frac{\sqrt3}{2}\\[1em] = \cancel{2} \times \frac{1}{\cancel{2}} \times \frac{\sqrt3}{2}\\[1em] = \frac{\sqrt3}{2}\end{aligned}$$

$$\begin{aligned}\text{3rd term} = \frac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \frac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \frac{2 \times \frac{1}{\sqrt3}}{1 + \Big(\frac{1}{\sqrt3}\Big)^2}\\[1em] = \frac{\frac{2}{\sqrt3}}{1 + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{3}{3} + \frac{1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{3 + 1}{3}}\\[1em] = \frac{\frac{2}{\sqrt3}}{\frac{4}{3}}\\[1em] = \frac{2 \times 3}{4 \times \sqrt3}\\[1em] = \frac{\sqrt3}{2}\end{aligned}$$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{sin 2 A} = \text{2 sin A cos A} = \frac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$.

$\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \frac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{cos 2 A} = \text{cos (2 x 30°)} = \text{cos 60°} = \frac{1}{2}$

$$\begin{aligned}\text{2nd term} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A}\\[1em] = \text{cos}^2 \text{ 30°} - \text{sin}^2 \text{ 30°}\\[1em] = \Big(\frac{\sqrt3}{2}\Big)^2 - \Big(\frac{1}{2}\Big)^2\\[1em] = \frac{3}{4} - \frac{1}{4}\\[1em] = \frac{3 - 1}{4}\\[1em] = \frac{2}{4}\\[1em] = \frac{1}{2}\end{aligned}$$

$$\begin{aligned}\text{3rd term} = \frac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \frac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \frac{1 - \Big(\frac{1}{\sqrt3}\Big)^2}{1 + \Big(\frac{1}{\sqrt3}\Big)^2}\\[1em] = \frac{\frac{3}{3} - \frac{1}{3}}{\frac{3}{3} + \frac{1}{3}}\\[1em] = \frac{\frac{3 - 1}{3}}{\frac{3 + 1}{3}}\\[1em] = \frac{\frac{2}{3}}{\frac{4}{3}}\\[1em] = \frac{\frac{2}{\cancel{3}}}{\frac{4}{\cancel{3}}}\\[1em] = \frac{2}{4}\\[1em] = \frac{1}{2}\end{aligned}$$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \frac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

2 cos² A - 1 = 1 - 2 sin² A

$$\begin{aligned}\text{L.H.S.} = 2 \text{cos}^2 A - 1\\[1em] = 2 \text{cos}^2 30° - 1\\[1em] = 2 \times \Big(\frac{\sqrt3}{2}\Big)^2 - 1\\[1em] = 2 \times \frac{3}{4} - 1\\[1em] = \frac{3}{2} - \frac{2}{2}\\[1em] = \frac{3 - 2}{2}\\[1em] = \frac{1}{2}\end{aligned}$$

$$\begin{aligned}\text{R.H.S.} = 1 - 2 \text{sin}^2 A \\[1em] = 1 - 2 \text{sin}^2 30°\\[1em] = 1 - 2 \times \Big(\frac{1}{2}\Big)^2\\[1em] = 1 - 2 \times \frac{1}{4}\\[1em] = \frac{2}{2} - \frac{1}{2}\\[1em] = \frac{2 - 1}{2}\\[1em] = \frac{1}{2}\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, 2 cos² A - 1 = 1 - 2 sin² A.

sin 3A = 3 sin A - 4 sin³ A

L.H.S. = sin 3A

= sin (3 x 30°)

= sin 90°

= 1

$$\begin{aligned}\text{R.H.S.} = 3 \text{sin A} - 4 \text{sin}^3 A\\[1em] = 3 \text{sin 30°} - 4 \text{sin}^3 30°\\[1em] = 3 \times \frac{1}{2} - 4 \times \Big(\frac{1}{2}\Big)^3\\[1em] = \frac{3}{2} - 4 \times \frac{1}{8}\\[1em] = \frac{3}{2} - \frac{1}{2}\\[1em] = \frac{3 - 1}{2}\\[1em] = \frac{2}{2}\\[1em] = 1\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, sin 3A = 3 sin A - 4 sin³ A.

sin (A - B) = sin A cos B - cos A sin B

L.H.S. = sin (A - B)

= sin (45° - 45°)

= sin 0°

= 0

R.H.S. = sin A cos B - cos A sin B

= sin 45° . cos 45° - cos 45°. sin 45°

$$\begin{aligned}= \frac {1}{\sqrt2} \times \frac {1}{\sqrt2} - \frac {1}{\sqrt2} \times \frac {1}{\sqrt2}\\[1em] = \frac {1}{2} - \frac {1}{2}\\[1em] = 0\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, sin (A - B) = sin A cos B - cos A sin B.

cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos (A + B)

= cos (45° + 45°)

= cos 90°

= 0

R.H.S. = cos A cos B - sin A sin B

= cos 45° . cos 45° - sin 45°. sin 45°

$$\begin{aligned}= \frac {1}{\sqrt2} \times \frac {1}{\sqrt2} - \frac {1}{\sqrt2} \times \frac {1}{\sqrt2}\\[1em] = \frac {1}{2} - \frac {1}{2}\\[1em] = 0\end{aligned}$$

∴ L.H.S. = R.H.S.

Hence, cos (A + B) = cos A cos B - sin A sin B.

2sin A - 1 = 0

⇒ 2sin A = 1

⇒ sin A = $\frac{1}{2}$

⇒ sin A = sin 30°

⇒ A = 30°

Hence, option 1 is the correct option.

cos A (cos A - 1) = 0

⇒ cos A = 0 or cos A = 1

⇒ cos A = cos 90° or cos A = cos 0°

The measure of angle A is 90° or 0°.

Hence, option 1 is the correct option.

tan⁴ A - 1 = 0

⇒ tan⁴A = 1

⇒ (tan A)⁴ = 1

⇒ tan A = tan 45°

⇒ A = 45°

Hence, option 2 is the correct option.

2sin 3A - 1 = 0

⇒ 2sin 3A = 1

⇒ sin 3A = $\frac{1}{2}$

⇒ sin 3A = sin 30°

So, 3A = 30°

⇒ A = $\frac{30°}{3}$

⇒ A = 10°

Hence, option 3 is the correct option.

3tan² A - 1 = 0

⇒ 3tan² A = 1

⇒ tan² A = $\frac{1}{3}$

⇒ tan A = $\sqrt{\frac{1}{3}}$

⇒ tan A = $\frac{1}{\sqrt3}$

⇒ tan A = tan 30°

⇒ A = 30°

Hence, option 4 is the correct option.

2 sin A = 1

⇒ sin A = $\frac{1}{2}$

⇒ sin A = sin 30°

Hence, A = 30°.

2 cos 2 A = 1

⇒ cos 2A = $\frac{1}{2}$

⇒ cos 2A = cos 60°

So, 2A = 60°

⇒ A = $\frac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

sin 3 A = $\frac{\sqrt3}{2}$

⇒ sin 3A = sin 60°

So, 3A = 60°

⇒ A = $\frac{60°}{3}$

⇒ A = 20°

Hence, A = 20°.

sec 2 A = 2

⇒ sec 2A = sec 60°

So, 2A = 60°

⇒ A = $\frac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

${\sqrt3}$ tan A = 1

⇒ tan A = $\frac{1}{\sqrt3}$

⇒ tan A = tan 30°

So, A = 30°

Hence, A = 30°.

tan 3 A = 1

⇒ tan 3A = tan 45°

So, 3A = 45°

⇒ A = $\frac{45°}{3}$

⇒ A = 15°

Hence, A = 15°.

2 sin 3 A = 1

⇒ sin 3A = $\frac{1}{2}$

⇒ sin 3A = sin 30°

So, 3A = 30°

⇒ A = $\frac{30°}{3}$

⇒ A = 10°

Hence, A = 10°.

${\sqrt3}$ cot 2 A = 1

⇒ cot 2A = $\frac{1}{\sqrt3}$

⇒ cot 2A = cot 60°

So, 2A = 60°

⇒ A = $\frac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

(sin A - 1) (2 cos A - 1) = 0

⇒ (sin A - 1) = 0 and (2 cos A - 1) = 0

⇒ sin A = 1 and cos A = $\frac{1}{2}$

⇒ sin A = sin 90° and cos A = cos 60°

Hence, A = 90° and 60°.

(tan A - 1) (cosec 3A - 1) = 0

⇒ (tan A - 1) = 0 and (cosec 3A - 1) = 0

⇒ tan A = 1 and cosec 3A = 1

⇒ tan A = tan 45° and cosec 3A = cosec 90°

So, A = 45° and 3A = 90°

⇒ A = 45° and A = 30°

Hence, A = 45° and 30°.

(sec 2A - 1) (cosec 3A - 1) = 0

⇒ sec 2A - 1 = 0 and cosec 3A - 1 = 0

⇒ sec 2A = 1 and cosec 3A = 1

⇒ sec 2A = sec 0° and cosec 3A = cosec 90°

So, 2A = 0° and 3A = 90°

⇒ A = 0° and A = $\frac{90°}{3} = 30°$

Hence, A = 0° and 30°.

cos 3A. (2 sin 2A - 1) = 0

⇒ cos 3A = 0 and 2sin 2A - 1 = 0

⇒ cos 3A = 0 and 2sin 2A = 1

⇒ cos 3A = 0 and sin 2A = $\frac{1}{2}$

⇒ cos 3A = cos 90° and sin 2A = sin 30°

So, 3A = 90° and 2A = 30°

⇒ A = $\frac{90°}{3} = 30°$ and A = $\frac{30°}{2} = 15°$

Hence, A = 30° and 15°.

(cosec 2A - 2) (cot 3A - 1) = 0

⇒ cosec 2A - 2 = 0 and cot 3A - 1 = 0

⇒ cosec 2A = 2 and cot 3A = 1

⇒ cosec 2A = cosec 30° and cot 3A = cot 45°

So, 2A = 30° and 3A = 45°

⇒ A = $\frac{30°}{2} = 15°$ and A = $\frac{45°}{3} = 15°$

Hence, A = 15°.

(i) 2 sin x° - 1 = 0

⇒ 2 sin x° = 1

⇒ sin x° = $\frac{1}{2}$

Hence, sin x° = $\frac{1}{2}$.

(ii) x°

⇒ sin x° = $\frac{1}{2}$

⇒ sin x° = sin 30°

Hence, x° = 30°.

(iii) cos x°

⇒ cos 30° = $\frac{\sqrt3}{2}$

Hence, cos x° = $\frac{\sqrt3}{2}$.

tan x°

⇒ tan 30° = $\frac{1}{\sqrt3}$

Hence, tan x° = $\frac{1}{\sqrt3}$.

(i) 4 cos² x° - 1 = 0

⇒ 4 cos² x° = 1

⇒ cos² x° = $\frac{1}{4}$

⇒ cos x° = $\sqrt{\frac{1}{4}}$

⇒ cos x° = $\frac{1}{2}$

⇒ cos x° = cos 60°

Hence, x° = 60°.

(ii) sin² x° + cos² x°

⇒ sin² 60° + cos² 60°

$$\begin{aligned}= \Big(\frac{\sqrt3}{2}\Big)^2 + \Big(\frac{1}{2}\Big)^2 \\[1em] = \frac{3}{4} + \frac{1}{4} \\[1em] = \frac{3 + 1}{4}\\[1em] = \frac{4}{4}\\[1em] = 1\end{aligned}$$

Hence, sin² x° + cos² x° = 1.

(iii) $\frac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

$$\begin{aligned}= \frac{1}{\text{cos}^2 \text{ 60°}} - \text{tan}^2 \text{ 60°}\\[1em] = \frac{1}{\Big(\frac{1}{2}\Big)^2} - (\sqrt3)^2\\[1em] = \frac{4}{1} - 3\\[1em] = 4 - 3\\[1em] = 1\end{aligned}$$

Hence, $\frac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°} = 1$.

4 sin² θ - 1 = 0

⇒ 4 sin² θ = 1

⇒ sin² θ = $\frac{1}{4}$

⇒ sin θ = $\sqrt{\frac{1}{4}}$

⇒ sin θ = $\frac{1}{2}$

⇒ sin θ = sin 30°

So, θ = 30°

Now, cos² θ + tan² θ

= cos² 30° + tan² 30°

$$\begin{aligned}= \Big(\frac{\sqrt3}{2}\Big)^2 + \Big(\frac{1}{\sqrt3}\Big)^2\\[1em] = \frac{3}{4} + \frac{1}{3}\\[1em] = \frac{3 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4}\\[1em] = \frac{9}{12} + \frac{4}{12}\\[1em] = \frac{9 + 4}{12}\\[1em] = \frac{13}{12}\\[1em] = 1\frac{1}{12}\end{aligned}$$

Hence, θ = 30° and cos² 30° + tan² 30° = $\frac{13}{12}$ = $1\frac{1}{12}$.

sin 3A = 1

⇒ sin 3A = sin 90°

So, 3A = 90°

⇒ A = $\frac{90°}{3} = 30°$

(i) sin A = sin 30° = $\frac{1}{2}$

Hence, sin A = $\frac{1}{2}$.

(ii) cos 2 A

= cos (2 x 30°)

= cos 60°

= $\frac{1}{2}$

Hence, cos 2A = $\frac{1}{2}$.

(iii) tan² A - $\frac{1}{\text{cos}^2 \text{ A}}$

$$\begin{aligned}= \text{tan}^2 \text{30°} - \frac{1}{\text{cos}^2 \text{30°}}\\[1em] = \Big(\frac{1}{\sqrt3}\Big)^2 - \frac{1}{\Big(\frac{\sqrt3}{2}\Big)^2}\\[1em] = \frac{1}{3} - \frac{1}{\frac{3}{4}}\\[1em] = \frac{1}{3} - \frac{4}{3}\\[1em] = \frac{1 - 4}{3}\\[1em] = \frac{- 3}{3}\\[1em] = - 1\end{aligned}$$

Hence, tan² A - $\frac{1}{\text{cos}^2 \text{ A}}$ = -1.

(i) 2 cos 2A = ${\sqrt3}$

⇒ cos 2A = $\frac{\sqrt3}{2}$

⇒ cos 2A = cos 30°

So, 2A = 30°

⇒ A = $\frac{30°}{2} = 15°$

Hence, A = 15°.

(ii) sin 3A

= sin (3 x 15°)

= sin 45°

= $\frac{1}{\sqrt2}$

Hence, sin 3A = $\frac{1}{\sqrt2}$.

(iii) sin² (75° - A) + cos² (45° + A)

= sin² (75° - 15°) + cos² (45° + 15°)

= sin² 60° + cos² 60°

$$\begin{aligned}= \Big(\frac{\sqrt3}{2}\Big)^2 + \Big(\frac{1}{2}\Big)^2\\[1em] = \frac{3}{4} + \frac{1}{4}\\[1em] = \frac{3 + 1}{4} \\[1em] = \frac{4}{4} \\[1em] = 1\end{aligned}$$

Hence, sin² (75° - A) + cos² (45° + A) = 1.

Given:

sin A = cos A

⇒ sin A = sin (90° - A)

So, A = 90° - A

⇒ A + A = 90°

⇒ 2A = 90°

⇒ A = $\frac{90°}{2}$

⇒ A = 45°

Hence, option 3 is the correct option.

Given:

sin A = cos B

⇒ sin A = sin (90° - B)

So, A = 90° - B

⇒ A + B = 90°

Hence, option 4 is the correct option.

Given:

$$\begin{aligned}\frac{\text{cos A}}{\text{sin B}}\times \frac{\text{tan B}}{\text{cot A}}\\[1em] ⇒ \frac{\text{cos A}}{\text{sin (90° - A)}}\times \frac{\text{tan (90° - A)}}{\text{cot A}}\\[1em] ⇒ \frac{\text{cos A}}{\text{cos A}}\times \frac{\text{cot A}}{\text{cot A}}\\[1em] ⇒ \frac{\cancel{cos A}}{\cancel{cos A}}\times \frac{\cancel{cot A}}{\cancel{cot A}}\\[1em] ⇒ 1\end{aligned}$$

Hence, option 1 is the correct option.

Given:

cosec 40° cos 50° + sin 50° sec 40°

= cosec 40° cos (90° - 40°) + sin (90° - 40°) sec 40°

= cosec 40° sin 40° + cos 40° sec 40°

= $\frac{1}{\text{sin 40°}} \times \text{sin 40°} + \text{cos 40°} \times \frac{1}{\text{cos 40°}}$

= 1 + 1

= 2

Hence, option 2 is the correct option.

Given:

In Δ ABC,

$$\begin{aligned}⇒ ∠ A + ∠ B + ∠ C = 180°\\[1em] ⇒ ∠ A + ∠ C = 180° - ∠ B\\[1em] ⇒ \frac{A + C}{2} = \frac{180° - B}{2}\\[1em] ⇒ \frac{A + C}{2} = 90° - \frac{B}{2}\\[1em] ⇒ \text{sec}\frac{A + C}{2} = \text{sec}{\Big(90° - \frac{B}{2}\Big)}\\[1em] ⇒ \text{sec}\frac{A + C}{2} = \text{cosec}\frac{B}{2}\\[1em]\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\frac{\text{cos 22°}}{\text{sin 68°}} = \frac{\text{cos (90° - 68°)}}{\text{sin 68°}}\\[1em] = \frac{\text{sin 68°}}{\text{sin 68°}}\\[1em] = \frac{\cancel{sin 68°}}{\cancel{sin 68°}}\\[1em] = 1\end{aligned}$$

Hence, $\frac{\text{cos 22°}}{\text{sin 68°}}$ = 1.

$$\begin{aligned}\frac{\text{tan 47°}}{\text{cot 43°}} = \frac{\text{tan (90° - 43°)}}{\text{cot 43°}}\\[1em] = \frac{\text{cot 43°}}{\text{cot 43°}}\\[1em] = \frac{\cancel{cot 43°}}{\cancel{cot 43°}}\\[1em] = 1\end{aligned}$$

Hence, $\frac{\text{tan 47°}}{\text{cot 43°}}$ = 1.

$$\begin{aligned}\frac{\text{sec 75°}}{\text{cosec 15°}} = \frac{\text{sec (90° - 15°)}}{\text{cosec 15°}}\\[1em] = \frac{\text{cosec 15°}}{\text{cosec 15°}}\\[1em] = \frac{\cancel{cosec 15°}}{\cancel{cosec 15°}}\\[1em] = 1\end{aligned}$$

Hence, $\frac{\text{sec 75°}}{\text{cosec 15°}}$ = 1.

$$\begin{aligned}\frac{\text{cos 55°}}{\text{sin 35°}} + \frac{\text{cot 35°}}{\text{tan 55°}}\\[1em] = \frac{\text{cos (90° - 35°)}}{\text{sin 35°}} + \frac{\text{cot (90° - 55°)}}{\text{tan 55°}}\\[1em] = \frac{\text{sin 35°}}{\text{sin 35°}} + \frac{\text{tan 55°}}{\text{tan 55°}}\\[1em] = \frac{\cancel{\sin 35°}}{\cancel{\sin 35°}} + \frac{\cancel{\tan 55°}}{\cancel{\tan 55°}}\\[1em] = 1 + 1\\[1em] = 2\end{aligned}$$

Hence, $\frac{\text{cos 55°}}{\text{sin 35°}} + \frac{\text{cot 35°}}{\text{tan 55°}}$ = 2.

sin² 40° - cos² 50°

= sin² (90° - 50°) - cos² 50°

= cos² 50° - cos² 50°

= 0

Hence, sin² 40° - cos² 50° = 0.

sec² 18° - cosec² 72°

= sec² (90° - 72°) - cosec² 72°

= cosec² 72° - cosec² 72°

= 0

Hence, sec² 18° - cosec² 72° = 0.

sin 15° cos 15° - cos 75° sin 75°

= sin (90° - 75°) cos (90° - 75°) - cos (90° - 15°) sin 75°

= cos 75° sin 75° - cos 75° sin 75°

= 0

Hence, sin 15° cos 15° - cos 75° sin 75° = 0.

sin 42° sin 48° - cos 42° cos 48°

= sin (90° - 48°) sin 48° - cos (90° - 48°) cos 48°

= cos 48° sin 48° - sin 48° cos 48°

= 0

Hence, sin 42° sin 48° - cos 42° cos 48° = 0.

sin (90° - A) sin A - cos (90° - A) cos A

= cos A sin A - sin A cos A

= 0

Hence, sin (90° - A) sin A - cos (90° - A) cos A = 0.

sin² 35° - cos² 55°

= sin² (90° - 55°) - cos² 55°

= cos² 55° - cos² 55°

= 0

Hence, sin² 35° - cos² 55° = 0.

$$\begin{aligned}\frac{\text{cot 54°}}{\text{tan 36°}} + \frac{\text{tan 20°}}{\text{cot 70°}} - 2\\[1em] = \frac{\text{cot (90° - 36°)}}{\text{tan 36°}} + \frac{\text{tan (90° - 70°)}}{\text{cot 70°}} - 2\\[1em] = \frac{\text{tan 36°}}{\text{tan 36°}} + \frac{\text{cot 70°}}{\text{cot 70°}} - 2\\[1em] = \frac{\cancel{tan 36°}}{\cancel{tan 36°}} + \frac{\cancel{cot 70°}}{\cancel{cot 70°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 2 - 2\\[1em] = 0\end{aligned}$$

Hence, $\frac{\text{cot 54°}}{\text{tan 36°}} + \frac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0$.

cos² 25° - sin² 65° - tan² 45°

= cos² (90° - 65°) - sin² 65° - tan² 45°

= sin² 65° - sin² 65° - tan² 45°

= - tan² 45°

= - 1

Hence, cos² 25° - sin² 65° - tan² 45° = -1.

$$\begin{aligned}\Big(\frac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\frac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\frac{\text{sin (90° - 13°)}}{\text{cos 13°}}\Big)^2 + \Big(\frac{\text{cos (90° - 13°)}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\frac{\text{cos 13°}}{\text{cos 13°}}\Big)^2 + \Big(\frac{\text{sin 13°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\frac{\cancel{cos 13°}}{\cancel{cos 13°}}\Big)^2 + \Big(\frac{\cancel{sin 13°}}{\cancel{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = 1^2 + 1^2 - 2 \times \Big(\frac{1}{\sqrt2}\Big)^2\\[1em] = 1 + 1 - 2 \times \Big(\frac{1}{2}\Big)\\[1em] = 2 - 1\\[1em] = 1\end{aligned}$$

Hence, $\Big(\frac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\frac{\text{cos 77°}}{\text{sin 13°}}\Big) - \text{2 cos}^2 45° = 1$.

tan 10° tan 15° tan 75° tan 80° = 1

L.H.S. = tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= $\frac{1}{\text{tan 80°}} \times \frac{1}{\text{tan 75°}} \times \text{tan 75°} \times \text{tan 80°}$

= $\frac{1}{\cancel{tan 80°}} \times \frac{1}{\cancel{tan 75°}} \times \cancel{tan 75°} \times \cancel{tan 80°}$

= 1

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, tan 10° tan 15° tan 75° tan 80° = 1.

sin 42° sec 48° + cos 42° cosec 48° = 2

L.H.S. = sin 42° sec 48° + cos 42° cosec 48°

= sin (90° - 48°) sec 48° + cos (90° - 48°) cosec 48°

= cos 48° sec 48° + sin 48° cosec 48°

= $\text{cos 48°} \frac{1}{\text{cos 48°}} + \text{sin 48°} \frac{1}{\text{sin 48°}}$

= $\cancel{cos 48°} \frac{1}{\cancel{cos 48°}} + \cancel{sin 48°} \frac{1}{\cancel{sin 48°}}$

= 1 + 1

= 2

R.H.S. = 2

∴ L.H.S. = R.H.S.

Hence, sin 42° sec 48° + cos 42° cosec 48° = 2.

sin 59° + tan 63°

= sin (90° - 31°) + tan (90° - 27°)

= cos 31° + cot 27°

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

cosec 68°+ cot 72°

= cosec (90° - 22°) + cot (90° - 18°)

= sec 22° + tan 18°

Hence, cosec 68°+ cot 72° = sec 22° + tan 18°.

cos 74° + sec 67°

= cos (90° - 16°) + sec (90° - 23°)

= sin 16° + cosec 23°

Hence, cos 74° + sec 67° = sin 16° + cosec 23°.

(i) $\text{sin}\frac{A+B}{2} = \text{cos}\frac{C}{2}$

According to angle sum property,

$$\begin{aligned}∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ A + ∠ B = 180° - ∠ C\\[1em] ⇒ \frac{A + B}{2} = \frac{180° - C}{2}\\[1em]\end{aligned}$$

$$\begin{aligned}\text{L.H.S.} = \text{sin}\frac{A+B}{2}\\[1em] = \text{sin}\frac{180° - C}{2}\\[1em] = \text{sin}\Big(90° - \frac{C}{2}\Big)\\[1em] = \text{cos}\frac{C}{2}\end{aligned}$$

R.H.S. = $\text{cos}\frac{C}{2}$

∴ L.H.S. = R.H.S.

Hence, $\text{sin}\frac{A+B}{2} = \text{cos}\frac{C}{2}$.

(ii) $\text{tan}\frac{B+C}{2} = \text{cot}\frac{A}{2}$

According to angle sum property,

$$\begin{aligned}∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ B + ∠ C = 180° - ∠ A\\[1em] ⇒ \frac{B + C}{2} = \frac{180° - A}{2}\\[1em]\end{aligned}$$

$$\begin{aligned}\text{L.H.S.} = \text{tan}\frac{B+C}{2}\\[1em] = \text{tan}\Big(\frac{180° - A}{2}\Big)\\[1em] = \text{tan}\Big(90° - \frac{A}{2}\Big)\\[1em] = \text{cot}\frac{A}{2}\end{aligned}$$

R.H.S. = $\text{cot}\frac{A}{2}$

∴ L.H.S. = R.H.S.

Hence, $\text{tan}\frac{B+C}{2} = \text{cot}\frac{A}{2}$.

$$\begin{aligned}\text{3}\frac{\text{ sin 72°}}{\text{ cos 18°}} - \frac{\text{sec 32°}}{\text{cosec 58°}}\\[1em] = \text{3}\frac{\text{ sin (90° - 18°)}}{\text{ cos 18°}} - \frac{\text{sec (90° - 58°)}}{\text{cosec 58°}}\\[1em] = \text{3}\frac{\text{ cos 18°}}{\text{ cos 18°}} - \frac{\text{cosec 58°}}{\text{cosec 58°}}\\[1em] = \text{3}\frac{\cancel{ cos 18°}}{\cancel{ cos 18°}} - \frac{\cancel{cosec 58°}}{\cancel{cosec 58°}}\\[1em] = 3 - 1\\[1em] = 2\end{aligned}$$

Hence, $\text{3}\frac{\text{ sin 72°}}{\text{ cos 18°}} - \frac{\text{sec 32°}}{\text{cosec 58°}} = 2$.

3 cos 80° cosec 10° + 2 sin 59° sec 31°

= 3 cos (90° - 10°) cosec 10° + 2 sin (90° - 31°) sec 31°

= 3 sin 10° cosec 10° + 2 cos 31° sec 31°

= $3 \text{sin 10°} \times \frac{1}{\text{sin 10°}} + 2 \text{cos 31°} \times \frac{1}{\text{cos 31°}}$

= $3 \cancel{sin 10°} \times \frac{1}{\cancel{sin 10°}} + 2 \cancel{cos 31°} \times \frac{1}{\cancel{cos 31°}}$

= 3 + 2

= 5

Hence, 3 cos 80° cosec 10° + 2 sin 59° sec 31° = 5.

$$\begin{aligned}\frac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°}\\[1em] = \frac{\text{sin (90° - 10°)}}{\text{cos 10°}} + \text{sin (90° - 31°) sec 31°}\\[1em] = \frac{\text{cos 10°}}{\text{cos 10°}} + \text{cos 31° sec 31°}\\[1em] = \frac{\cancel{cos 10°}}{\cancel{cos 10°}} + \text{cos 31°}\frac{1}{\text{cos 31°}}\\[1em] = 1 + \cancel{cos 31°}\frac{1}{\cancel{cos 31°}}\\[1em] = 1 + 1\\[1em] = 2\end{aligned}$$

Hence, $\frac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°} = 2$.

tan (55° - A) - cot (35° + A)

= tan [(90° - 35°) - A] - cot (35° + A)

= tan [90° - (35° + A)] - cot (35° + A)

= cot (35° + A) - cot (35° + A)

= 0

Hence, tan (55° - A) - cot (35° + A) = 0.

cosec (65° + A) - sec (25° - A)

= cosec [(90° - 25°) + A] - sec (25° - A)

= cosec [90° - (25° - A)] - sec (25° - A)

= sec (25° - A) - sec (25° - A)

= 0

Hence, cosec (65° + A) - sec (25° - A) = 0.

$$\begin{aligned}2\frac{\text{tan 57°}}{\text{cot 33°}} - \frac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\frac{\text{tan (90° - 33°)}}{\text{cot 33°}} - \frac{\text{cot (90° - 20°)}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\frac{\text{cot 33°}}{\text{cot 33°}} - \frac{\text{tan 20°}}{\text{tan 20°}} - {\sqrt2} \frac{1}{\sqrt2}\\[1em] = 2\frac{\cancel{cot 33°}}{\cancel{cot 33°}} - \frac{\cancel{tan 20°}}{\cancel{tan 20°}} - \cancel{\sqrt2} \frac{1}{\cancel{\sqrt2}}\\[1em] = 2 - 1 - 1\\[1em] = 0\end{aligned}$$

Hence, $2\frac{\text{tan 57°}}{\text{cot 33°}} - \frac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°} = 0$.

$$\begin{aligned}\frac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\frac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \frac{\text{cot}^2 \text{ (90° - 49°)}}{\text{tan}^2 \text{ 49°}} - 2\frac{\text{ sin}^2 \text{ (90° - 15°)}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \frac{\text{tan}^2 \text{49°}}{\text{tan}^2 \text{ 49°}} - 2\frac{\text{ cos}^2 \text{15°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \frac{\cancel{tan^2 49°}}{\cancel{tan^2 49°}} - 2\frac{\cancel{ cos^2 15°}}{\cancel{ cos^2 15°}}\\[1em] = 1 - 2\\[1em] = - 1\end{aligned}$$

Hence, $\frac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\frac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}} = -1$.

$$\begin{aligned}\frac{\text{cos 70°}}{\text{sin 20°}} + \frac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°}\\[1em] =\frac{\text{cos (90° - 70°)}}{\text{sin 20°}} + \frac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8 \times \Big(\frac{1}{2}\Big)^2\\[1em] =\frac{\text{sin 70°}}{\text{sin 20°}} + \frac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\frac{1}{4}\Big)\\[1em] =\frac{\cancel{sin 70°}}{\cancel{sin 20°}} + \frac{\cancel{sin 31°}}{\cancel{sin 31°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 0\end{aligned}$$

Hence, $\frac{\text{cos 70°}}{\text{sin 20°}} + \frac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°} = 0$.

14 sin 30° + 6 cos 60° - 5 tan 45°

$= 14 \times \frac{1}{2} + 6 \times \frac{1}{2} - 5 \times 1$

= 7 + 3 - 5

= 10 - 5

= 5

Hence, 14 sin 30° + 6 cos 60° - 5 tan 45° = 5.

Given:

ABC is right-angled triangle at B.

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 90° + ∠ C = 180°

⇒ ∠ A + ∠ C = 180° - 90°

⇒ ∠ A + ∠ C = 90°

⇒ ∠ A = 90° - ∠ C

Now,

$$\begin{aligned}\frac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}\\[1em] = \frac{\text{sec (90° - C) . sin C - tan (90° - C). tan C}}{\text{sin B}}\\[1em] = \frac{\text{cosec C . sin C - cot C. tan C}}{\text{sin B}}\\[1em] = \frac{\frac{1}{\text{sin C}} \times \text{sin C} - \frac{1}{\text{tan C}} \times \text{tan C}}{\text{sin B}}\\[1em] = \frac{\frac{1}{\cancel{sin C}} \times \cancel{sin C} - \frac{1}{\cancel{tan C}} \times \cancel{tan C}}{\text{sin B}}\\[1em] = \frac{1 - 1}{\text{sin B}}\\[1em] = 0\end{aligned}$$

Hence, $\frac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}} = 0$.

(i) sin (90° - 3A) . cosec 42° = 1

⇒ cos 3A . cosec 42° = 1

⇒ cos 3A x $\frac{1}{\text{sin 42°}}$ = 1

⇒ cos 3A = sin 42°

⇒ cos 3A = sin (90° - 48°)

⇒ cos 3A = cos 48°

So, 3A = 48°

⇒ A = $\frac{48°}{3}$

⇒ A = 16°

Hence, A = 16°.

(ii) cos (90° - A) . sec 77° = 1

⇒ sin A . sec 77° = 1

⇒ sin A x $\frac{1}{\text{cos 77°}}$ = 1

⇒ sin A = cos 77°

⇒ sin A = cos (90° - 13°)

⇒ sin A = sin 13°

So, A = 13°

Hence, A = 13°.

Since ΔPQR is a right angled triangle, using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ PQ² = QR² + PR²

⇒ 13² = PR² + 12²

⇒ 169 = PR² + 144

⇒ PR² = 169 - 144

⇒ PR² = 25

⇒ PR = $\sqrt{25}$

⇒ PR = 5 cm.

By formula, tan A = $\frac{\text{Height}}{\text{Base}}$

From figure,

$$\begin{aligned}\text{tan A} = \frac{PR}{RQ} \\[1em] = \frac{5}{12}.\end{aligned}$$

Hence, option 1 is the correct option.

Since ΔABC is a right angled triangle.

By formula,

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ and cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$.

So, sin B = $\frac{\text{AC}}{\text{AB}} = \frac{3}{5}$

cos A = $\frac{\text{AC}}{\text{AB}} = \frac{3}{5}$

The value of sin B - cos A = $\frac{3}{5} - \frac{3}{5}$ = 0.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \text{cos 60° - sin 90° + 2 cos 0° } = \frac{1}{2} - 1 + 2 \times 1\\[1em] = \frac{1}{2} - 1 + 2\\[1em] = \frac{1}{2} + 1\\[1em] = \frac{1 + 2}{2}\\[1em] = \frac{3}{2}\\[1em] = 1\frac{1}{2}.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

⇒ sin 23° - cos 67°

⇒ sin 23° - cos (90° - 23°)

⇒ sin 23° - sin 23°

⇒ 0.

Hence, option 2 is the correct option.

Given, ∠C = 90°.

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° - 90°

⇒ ∠A + ∠B = 90°

⇒ ∠A = 90° - ∠B

So, statement 2 is true.

⇒ tan A = tan (90° - ∠B)

⇒ tan A = cot B

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given,

⇒ 4 cos A = 3

⇒ cos A = $\frac{3}{4}$

⇒ $\frac{1}{\text{cos A}} = \frac{1}{\frac{3}{4}}$

⇒ sec A = $\frac{4}{3}$.

∴ Both the statements are true.

Hence, option 1 is the correct option.

We know that,

sin 30° = $\frac{1}{2}$, cos 60° = $\frac{1}{2}$, and tan 45° = 1.

So, reason (R) is true.

$$\begin{aligned}\text{sin}^2 30° - \text{2 cos}^3 60° + \text{2 tan}^4 45° = \Big(\frac{1}{2}\Big)^2 - 2 \times \Big(\frac{1}{2}\Big)^3 + 2 \times 1^4\\[1em] = \frac{1}{4} - 2 \times \frac{1}{8} + 2\\[1em] = \frac{1}{4} - \frac{1}{4} + 2\\[1em] = 2.\end{aligned}$$

So, assertion (A) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

If A = 30°,

60° - A = 60° - 30° = 30° and 60° + A = 60° + 30° = 90°.

So, reason (R) is true.

sin A = sin 30° = $\frac{1}{2}$

sin (60° - A) = sin 30° = $\frac{1}{2}$

sin (60° + A) = sin 90° = 1

Substituting values in 4 sin A sin (60° - A) sin (60° + A), we get :

$$\begin{aligned}\Rightarrow 4 \times \frac{1}{2} \times \frac{1}{2} \times 1\\[1em] \Rightarrow 4 \times \frac{1}{4}\\[1em] \Rightarrow 1.\end{aligned}$$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Given:

$\text{cosec θ} = {\sqrt5}$

$$\begin{aligned}⇒ \text{cosec θ} = \frac{Hypotenuse}{Perpendicular} = {\sqrt5}\\[1em]\end{aligned}$$

∴ If length of AC = $\sqrt{5}$x unit, length of BC = x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ ($\sqrt{5}$x)² = (x)² + AB²

⇒ 5x² = x² + AB²

⇒ AB² = 5x² - x²

⇒ AB² = 4x²

⇒ AB = $\sqrt{4 \text{x}^2}$

⇒ AB = 2x

(i) sin θ = $\frac{Perpendicular}{Hypotenuse}$

$= \frac{BC}{CA} = \frac{x}{\sqrt{5}x} = \frac{1}{\sqrt{5}}$

cos θ = $\frac{Base}{Hypotenuse}$

$= \frac{AB}{AC} = \frac{2x}{\sqrt{5}x} = \frac{2}{\sqrt{5}}$

Now,

2 - sin²θ - cos²θ

$$\begin{aligned}= 2 - \Big(\frac{1}{\sqrt{5}}\Big)^2 - \Big(\frac{2}{\sqrt{5}}\Big)^2\\[1em] = 2 - \frac{1}{5} - \frac{4}{5}\\[1em] = 2 + \frac{-1 - 4}{5}\\[1em] = 2 + \frac{-5}{5}\\[1em] = 2 - 1\\[1em] = 1\end{aligned}$$

Hence, 2 - sin²θ - cos²θ = 1.

(ii) $2 + \frac{1}{\text{sin}^2 \text{ θ}} - \frac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$

$$\begin{aligned}= 2 + \frac{1}{\Big(\frac{1}{\sqrt{5}}\Big)^2} - \frac{\Big(\frac{2}{\sqrt{5}}\Big)^2}{\Big(\frac{1}{\sqrt{5}}\Big)^2}\\[1em] = 2 + \frac{1}{\frac{1}{5}} - \frac{\frac{4}{5}}{\frac{1}{5}}\\[1em] = 2 + \frac{5}{1} - \frac{\frac{4}{\cancel{5}}}{\frac{1}{\cancel{5}}}\\[1em] = 2 + 5 - \frac{4}{1}\\[1em] = 7 - 4\\[1em] = 3\end{aligned}$$

Hence, $2 + \frac{1}{\text{sin}^2 \text{ θ}} - \frac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$ = 3.

Since D is the mid-point of A. So, AC = 2DC

(i) $\text{tan ∠CAB} = \frac{Perpendicular}{Base} = \frac{BC}{AC}$

$\text{tan ∠CDB} = \frac{Perpendicular}{Base} = \frac{BC}{DC}$

Now,

$$\begin{aligned}\frac{\text{tan ∠CAB}}{\text{tan ∠CDB}}\\[1em] = \frac{\frac{BC}{AC}}{\frac{BC}{DC}}\\[1em] = \frac{BC \times DC}{AC \times BC}\\[1em] = \frac{\cancel{BC} \times DC}{AC \times \cancel{BC}}\\[1em] = \frac{DC}{AC}\\[1em] = \frac{DC}{2 \times DC}\\[1em] = \frac{\cancel{DC}}{2 \times \cancel{DC}}\\[1em] = \frac{1}{2}\end{aligned}$$

Hence, $\frac{\text{tan ∠CAB}}{\text{tan ∠CDB}} = \frac{1}{2}$

(ii) $\text{tan ∠ABC} = \frac{Perpendicular}{Base} = \frac{AC}{BC}$

$\text{tan ∠DBC} = \frac{Perpendicular}{Base} = \frac{DC}{BC}$

Now,

$$\begin{aligned}\frac{\text {tan ∠ABC}}{\text {tan ∠DBC}}\\[1em] = \frac{\frac{AC}{BC}}{\frac{DC}{BC}}\\[1em] = \frac{AC \times BC}{BC \times DC}\\[1em] = \frac{AC \times \cancel{BC}}{\cancel{BC} \times DC}\\[1em] = \frac{AC}{DC}\\[1em] = \frac{2 \times DC}{DC}\\[1em] = \frac{2 \times \cancel{DC}}{\cancel{DC}}\\[1em] = 2\end{aligned}$$

Hence, $\frac{\text {tan ∠ABC}}{\text {tan ∠DBC}} = 2$

Given:

3 cos A = 4 sin A

$\frac{\text{cos A}}{\text{sin A}} = \frac{4}{3}$

$\text{cot A} = \frac{4}{3}$

$\text{cot A} = \frac{Base}{Perpendicular} = \frac{4}{3}$

∴ If length of AB = 4x unit, length of BC = 3x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (4x)² + (3x)²

⇒ AC² = 16x² + 9x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

(i) cos $A = \frac{Base}{Hypotenuse}$

$= \frac{AB}{AC} = \frac{4x}{5x} = \frac{4}{5}$

Hence, cos $A = \frac{4}{5}$.

(ii) 3 - cot² A + cosec² A

cot $A = \frac{Base}{Perpendicular}$

$= \frac{AB}{BC} = \frac{4x}{3x} = \frac{4}{3}$

cosec $A = \frac{Hypotenuse}{Perpendicular}$

$= \frac{AC}{BC} = \frac{5x}{3x} = \frac{5}{3}$

Now,

3 - cot² A + cosec² A

$$\begin{aligned}= 3 - \Big(\frac{4}{3}\Big)^2 + \Big(\frac{5}{3}\Big)^2\\[1em] = 3 - \frac{16}{9} + \frac{25}{9}\\[1em] = 3 + \frac{-16 + 25}{9}\\[1em] = 3 + \frac{9}{9}\\[1em] = 3 + 1\\[1em] = 4\end{aligned}$$

Hence, 3 - cot² A + cosec² A = 4.

From the figure, in Δ ADC,

⇒ AC² = DC² + AD² (∵ AC is hypotenuse)

⇒ 20² = DC² + 12²

⇒ 400 = DC² + 144

⇒ DC² = 400 - 144

⇒ DC² = 256

⇒ DC = $\sqrt{256}$

⇒ DC = 16

BD = BC - DC

= 21 - 16 = 5

In Δ ABD,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = 12² + 5²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$

⇒ AB = 13

sin x = $\frac{Perpendicular}{Hypotenuse}$

$= \frac{BD}{AB} = \frac{5}{13}$

sin y = $\frac{Perpendicular}{Hypotenuse}$

$= \frac{AD}{AC} = \frac{12}{20} = \frac{3}{5}$

cot y = $\frac{Base}{Perpendicular}$

$= \frac{DC}{AD} = \frac{16}{12} = \frac{4}{3}$

Now,

$$\begin{aligned}\frac{10}{\text{sin x}} + \frac{6}{\text{sin y}} - \text{6 cot y}\\[1em] = \frac{10}{\frac{5}{13}} + \frac{6}{\frac{3}{5}} - 6 \times \frac{4}{3}\\[1em] = \frac{10 \times 13}{5} + \frac{6 \times 5}{3} - \frac{24}{3}\\[1em] = \frac{130}{5} + \frac{30}{3} - \frac{24}{3}\\[1em] = 26 + 10 - 8\\[1em] = 28\end{aligned}$$

Hence, $\frac{10}{\text{sin x}} + \frac{6}{\text{sin y}} - \text{6 cot y} = 28.$

$$\begin{aligned}\frac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}\\[1em] = \frac{\text{cos (3 x 15°) - 2 cos (4 x 15°)}}{\text{sin (3 x 15°) + 2 sin (4 x 15°)}}\\[1em] = \frac{\text{cos 45° - 2 cos 60°}}{\text{sin 45° + 2 sin 60°}}\\[1em] = \frac{\Big(\frac{1}{\sqrt2}\Big) - 2 \times \Big(\frac{1}{2}\Big)}{\Big(\frac{1}{\sqrt2}\Big) + 2 \times \Big(\frac{\sqrt3}{2}\Big)}\\[1em] = \frac{\frac{1}{\sqrt2} - 1}{\frac{1}{\sqrt2} + \sqrt3}\\[1em] = \frac{\frac{1}{\sqrt2} - \frac{1 \times \sqrt2}{\sqrt2}}{\frac{1}{\sqrt2} + \frac{\sqrt3 \times \sqrt2}{\sqrt2}}\\[1em] = \frac{\frac{1}{\sqrt2} - \frac{\sqrt2}{\sqrt2}}{\frac{1}{\sqrt2} + \frac{\sqrt6}{\sqrt2}}\\[1em] = \frac{\frac{1 - \sqrt2}{\sqrt2}}{\frac{1 + \sqrt6}{\sqrt2}}\\[1em] = \frac{\frac{1 - \sqrt2}{\cancel{\sqrt2}}}{\frac{1 + \sqrt6}{\cancel{\sqrt2}}}\\[1em] = \frac{1 - \sqrt2}{1 + \sqrt6}\\[1em] = \frac{(1 - \sqrt2) \times (1 - \sqrt6)}{(1 + \sqrt6) \times (1 - \sqrt6)}\\[1em] = \frac{1 - \sqrt2 - \sqrt6 + \sqrt{12}}{1 - 6}\\[1em] = \frac{1 - \sqrt2 - \sqrt6 + 2\sqrt3}{-5}\\[1em] = \frac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)\end{aligned}$$

Hence, $\frac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}} = \frac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

$$\begin{aligned}\frac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}\\[1em] = \frac{\text{3 sin (3 x 20°) + 2 cos((2 x 20°) + 5°)}}{\text{2 cos (3 x 20°) - sin((2 x 20°) - 10°)}}\\[1em] = \frac{\text{3 sin 60° + 2 cos(40° + 5°)}}{\text{2 cos 60° - sin(40° - 10°)}}\\[1em] = \frac{\text{3 sin 60° + 2 cos 45°}}{\text{2 cos 60° - sin 30°}}\\[1em] = \frac{3 \times \frac{\sqrt3}{2} + 2 \times \frac{1}{\sqrt2}}{2 \times \frac{1}{2} - \frac{1}{2}}\\[1em] = \frac{\frac{3\sqrt3}{2} + \frac{2}{\sqrt2}}{\frac{2}{2} - \frac{1}{2}}\\[1em] = \frac{\frac{3\sqrt3}{2} + \frac{2 \times \sqrt2}{\sqrt2 \times \sqrt2}}{\frac{2 - 1}{2}}\\[1em] = \frac{\frac{3\sqrt3}{2} + \frac{2\sqrt2}{2}}{\frac{1}{2}}\\[1em] = \frac{\frac{3\sqrt3 + 2\sqrt2}{2}}{\frac{1}{2}}\\[1em] = \frac{\frac{3\sqrt3 + 2\sqrt2}{\cancel2}}{\frac{1}{\cancel2}}\\[1em] = 3\sqrt3 + 2\sqrt2\end{aligned}$$

Hence, $\frac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}} = 3\sqrt3 + 2\sqrt2.$

2 cos 3x - 1 = 0

⇒ 2 cos 3x = 1

⇒ cos 3x = $\frac{1}{2}$

⇒ cos 3x = cos 60°

So, 3x = 60°

⇒ x = $\frac{60°}{3} = 20°$

Hence, x = 20°.

$\text{cos }\frac{x}{3} - 1 = 0$

$$\begin{aligned}⇒ \text{cos } \frac{x}{3} = 1\\[1em] ⇒ \text{cos } \frac{x}{3} = \text{cos 0°}\end{aligned}$$

So,

$$\begin{aligned}⇒ \frac{x}{3} = 0°\\[1em] ⇒ x = 3 \times 0°\\[1em] ⇒ x = 0°\\[1em]\end{aligned}$$

Hence, x = 0°.

sin (x + 10°) = $\frac{1}{2}$

⇒ sin (x + 10°) = sin 30°

So, x + 10° = 30°

⇒ x = 30° - 10° = 20°

Hence, x = 20°.

cos (2x - 30°) = 0

⇒ cos (2x - 30°) = cos 90°

So, 2x - 30° = 90°

⇒ 2x = 90° + 30°

⇒ x = $\frac{120°}{2}$

⇒ x = 60°

Hence, x = 60°.

2 cos (3x - 15°) = 1

⇒ cos (3x - 15°) = $\frac{1}{2}$

⇒ cos (3x - 15°) = cos 60°

So, 3x - 15° = 60°

⇒ 3x = 60° + 15°

⇒ x = $\frac{75°}{3}$

⇒ x = 25°

Hence, x = 25°.

tan² (x - 5°) = 3

⇒ tan (x - 5°) = $\sqrt3$

⇒ tan (x - 5°) = tan 60°

So, x - 5° = 60°

⇒ x = 60° + 5°

⇒ x = 65°

Hence, x = 65°.

3 tan² (2x - 20°) = 1

⇒ tan² (2x - 20°) = $\frac{1}{3}$

⇒ tan (2x - 20°) = $\sqrt\frac{1}{3}$

⇒ tan (2x - 20°) = $\frac{1}{\sqrt3}$

⇒ tan (2x - 20°) = tan 30°

So, 2x - 20° = 30°

⇒ 2x = 30° + 20°

⇒ 2x = 50°

⇒ x = $\frac{50°}{2}$

⇒ x = 25°

Hence, x = 25°.

$\text{cos}\Big(\frac{x}{2} + 10°\Big)$ = $\frac{\sqrt3}{2}$

$⇒ \text{cos } \Big(\frac{x}{2} + 10°\Big) = \text{cos 30°}$

So,

$$\begin{aligned}⇒ \Big(\frac{x}{2} + 10°\Big) = 30°\\[1em] ⇒ \frac{x}{2} = 30° - 10°\\[1em] ⇒ x = 20° \times 2\\[1em] ⇒ x = 40° \\[1em]\end{aligned}$$

Hence, x = 40°.

sin² x + sin² 30° = 1

$$\begin{aligned}⇒ \text{sin }^2 x + \Big(\frac{1}{2}\Big)^2 = 1\\[1em] ⇒ \text{sin }^2 x + \Big(\frac{1}{4}\Big) = 1\\[1em] ⇒ \text{sin }^2 x = 1 - \Big(\frac{1}{4}\Big)\\[1em] ⇒ \text{sin }^2 x = \frac{4}{4} - \frac{1}{4}\\[1em] ⇒ \text{sin }^2 x = \frac{4 - 1}{4}\\[1em] ⇒ \text{sin }^2 x = \frac{3}{4}\\[1em] ⇒ \text{sin } x = \sqrt\frac{3}{4}\\[1em] ⇒ \text{sin } x = \frac{\sqrt3}{2}\\[1em] ⇒ \text{sin } x = \text{sin } 60°\\[1em]\end{aligned}$$

So, x = 60°

Hence, x = 60°.

cos² 30° + cos² x = 1

$$\begin{aligned}⇒ \Big(\frac{\sqrt{3}}{2}\Big)^2 + \text{cos }^2 x = 1\\[1em] ⇒ \Big(\frac{3}{4}\Big) + \text{cos }^2 x = 1\\[1em] ⇒ \text{cos }^2 x = 1 - \Big(\frac{3}{4}\Big)\\[1em] ⇒ \text{cos }^2 x = \frac{4}{4} - \frac{3}{4}\\[1em] ⇒ \text{cos }^2 x = \frac{4 - 3}{4}\\[1em] ⇒ \text{cos }^2 x = \frac{1}{4}\\[1em] ⇒ \text{cos } x = \sqrt\frac{1}{4}\\[1em] ⇒ \text{cos } x = \frac{1}{2}\\[1em] ⇒ \text{cos } x = \text{cos 60°} \\[1em]\end{aligned}$$

So, x = 60°

Hence, x = 60°.

cos² 30° + sin² 2x = 1

$$\begin{aligned}⇒ \Big(\frac{\sqrt3}{2}\Big)^2 + \text{sin }^2 2x = 1\\[1em] ⇒ \Big(\frac{3}{4}\Big) + \text{sin }^2 2x = 1\\[1em] ⇒ \text{sin}^2 2x = 1 - \frac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \frac{4}{4} - \frac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \frac{4 - 3}{4}\\[1em] ⇒ \text{sin}^2 2x = \frac{1}{4}\\[1em] ⇒ \text{sin} 2x = \sqrt\frac{1}{4}\\[1em] ⇒ \text{sin} 2x = \frac{1}{2}\\[1em] ⇒ \text{sin} 2x = \text{sin }30°\end{aligned}$$

So, 2x = 30°

⇒ x = $\frac{30°}{2}$

⇒ x = 15°

Hence, x = 15°.

sin² 60° + cos² (3x - 9°) = 1

$$\begin{aligned}⇒ \Big(\frac{\sqrt3}{2}\Big)^2 + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \Big(\frac{3}{4}\Big) + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \text{cos}^2 (3x - 9°) = 1 - \frac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \frac{4}{4} - \frac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \frac{4 - 3}{4} \\[1em] ⇒ \text{cos}^2 (3x - 9°) = \frac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \sqrt\frac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \frac{1}{2} \\[1em] ⇒ \text{cos} (3x - 9°) = \text{cos } 60°\end{aligned}$$

So, 3x - 9° = 60°

⇒ 3x = 60° + 9°

⇒ 3x = 69°

⇒ x = $\frac{69°}{3}$

⇒ x = 23°

Hence, x = 23°.

2 cos (A + B) = 1

⇒ cos (A + B) = $\frac{1}{2}$

⇒ cos (A + B) = cos 60°

So, A + B = 60° ...............(1)

2 sin (A - B) = 1

⇒ sin (A - B) = $\frac{1}{2}$

⇒ sin (A - B) = sin 30°

So, A - B = 30° ...............(2)

Adding equation (1) and (2), we get

(A + B) + (A - B) = 60° + 30°

⇒ A + B + A - B = 90°

⇒ 2A = 90°

⇒ A = $\frac{90°}{2}$

⇒ A = 45°

From equation (2), A - B = 30°

⇒ 45° - B = 30°

⇒ B = 45° - 30°

⇒ B = 15°

Hence, A = 45° and B = 15°.

For triangle ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ B + ∠ C = 180° - ∠ A

⇒ $\frac{B + C}{2} = \frac{180° - A}{2}$

⇒ $\frac{B + C}{2} = 90° - \frac{A}{2}$

$$\begin{aligned}\text{L.H.S.} = \text{sin}^2\frac{A}{2} + \text{sin}^2\frac{B+C}{2}\\[1em] = \text{sin}^2\frac{A}{2} + \text{sin}^2\Big(90° - \frac{A}{2}\Big)\\[1em] = \text{sin}^2\frac{A}{2} + \text{cos}^2\frac{A}{2}\\[1em] = 1\end{aligned}$$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, $\text{sin}^2\frac{A}{2}$ + $\text{sin}^2\frac{B+C}{2}$ = 1.

Given:

sec (90° - 3A) . cos 48° = 1

⇒ cosec 3A . cos 48° = 1

⇒ $\frac{1}{\text{sin 3A}}$ . cos 48° = 1

⇒ sin 3A = cos 48°

⇒ sin 3A = cos (90° - 42°)

⇒ sin 3A = sin 42°

So, 3A = 42°

⇒ A = $\frac{42°}{3}$

⇒ A = 14°

Hence, A = 14°.

In △ ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + ∠ B + 90° = 180°

⇒ ∠ A + ∠ B = 180° - 90°

⇒ ∠ A + ∠ B = 90°

⇒ sin(A + B) = sin 90°

⇒ sin(A + B) = 1

Hence, sin (A + B) = 1.

sec A sin A = 0

⇒ $\frac{1}{\text{cos A}}$ sin A = 0

⇒ tan A = 0

⇒ tan A = tan 0°

Thus, A = 0°.

Now, cos A = cos 0° = 1

Hence, cos A = 1.

We know that,

sec θ = cosec (90° - θ)

Solving,

⇒ sec 2A = cosec (A + 48°)

⇒ cosec (90° - 2A) = cosec (A + 48°)

⇒ 90° - 2A = A + 48°

⇒ A + 2A = 90° - 48°

⇒ 3A = 42°

⇒ A = $\frac{42°}{3}$ = 14°.

Hence, A = 14°.