Trigonometrical Ratios of Standard Angles

(i) 7 sin 30° cos 60°

$$\begin{aligned}= 7 \times \frac{1}{2} \times \frac{1}{2} \\[1em] = \frac{7}{4}.\end{aligned}$$

Hence, 7 sin 30° cos 60° = $\frac{7}{4}$.

(ii) 3 sin² 45° + 2 cos² 60°

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

Hence, 3 sin² 45° + 2 cos² 60° = 2.

(iii) cos² 45° + sin² 60° + sin² 30°

$$\begin{aligned}= \Big(\frac{1}{\sqrt{2}}\Big)^2 + \Big(\frac{\sqrt{3}}{2}\Big)^2 + \Big(\frac{1}{2}\Big)^2 \\[1em] = \frac{1}{2} + \frac{3}{4} + \frac{1}{4} \\[1em] = \frac{2 + 3 + 1}{4} \\[1em] = \frac{6}{4} \\[1em] = \frac{3}{2}.\end{aligned}$$

Hence, cos² 45° + sin² 60° + sin² 30° = $\frac{3}{2}$.

(iv) cos 90° + cos² 45° sin 30° tan 45°

$$\begin{aligned}= 0 + \Big(\frac{1}{\sqrt{2}}\Big)^2 \times \frac{1}{2} \times 1 \\[1em] = 0 + \frac{1}{2} \times \frac{1}{2} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, cos 90° + cos² 45° sin 30° tan 45° = $\frac{1}{4}$.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 45° + \text{cos}^2 45°}{\text{tan}^2 60°} \\[1em] [\because \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} = 1], \\[1em] \Rightarrow \frac{1}{(\sqrt{3})^2} \\[1em] \Rightarrow \frac{1}{3}\end{aligned}$$

Hence, $\frac{\text{sin}^2 45° + \text{cos}^2 45°}{\text{tan}^2 60°} = \frac{1}{3}$.

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin 30° - sin 90° + 2 cos 0°}}{\text{tan 30° × tan 60°}} \\[1em] \Rightarrow \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} \\[1em] \Rightarrow \frac{\frac{1 - 2 + 4}{2}}{1} \\[1em] \Rightarrow \frac{3}{2}.\end{aligned}$$

Hence, $\frac{\text{sin 30° - sin 90° + 2 cos 0°}}{\text{tan 30° × tan 60°}} = \frac{3}{2}$.

(iii) Solving,

$$\begin{aligned}\Rightarrow\frac{4}{3} \text{tan}^2 30° + \text{sin}^2 60° - \text{3 cos}^2 60° + \frac{3}{4}\text{tan}^2 60° - 2\text{tan}^2 45° \\[1em] \Rightarrow \frac{4}{3} \times \Big(\frac{1}{\sqrt{3}}\Big)^2 + \Big(\frac{\sqrt{3}}{2}\Big)^2 - 3 \times \Big(\frac{1}{2}\Big)^2 + \frac{3}{4} \times (\sqrt{3})^2 - 2 \times (1)^2 \\[1em] \Rightarrow \frac{4}{3} \times \frac{1}{3} + \frac{3}{4} - \frac{3}{4} + \frac{3}{4} \times 3 - 2 \\[1em] \Rightarrow \frac{4}{9} + \frac{9}{4} - 2 \\[1em] \Rightarrow \frac{16 + 81 - 72}{36} \\[1em] \Rightarrow \frac{25}{36}.\end{aligned}$$

Hence, $\frac{4}{3} \text{tan}^2 30° + \text{sin}^2 60° - \text{3 cos}^2 60° + \frac{3}{4}\text{tan}^2 60° - 2\text{tan}^2 45° = \frac{25}{36}$.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{\frac{\sqrt{3}}{2}}{\Big(\frac{1}{\sqrt{2}}\Big)^2} - 3 \times \frac{1}{\sqrt{3}} + 5 \times 0 \\[1em] \Rightarrow \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} - \sqrt{3} \\[1em] \Rightarrow \frac{2\sqrt{3}}{2} - \sqrt{3} \\[1em] \Rightarrow \sqrt{3} - \sqrt{3} \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $\frac{\text{sin 60°}}{\text{cos}^2 45°} - 3\text{ tan 30° + 5 cos 90°} = 0.$

(ii) Solving,

$$\begin{aligned}\Rightarrow 2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°} \\[1em] \Rightarrow 2\sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2} + 2 \times \sqrt{3} \times \frac{1}{2} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + \sqrt{3} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + 3 - 1 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°} = 3.$

(iii) Solving,

$$\begin{aligned}\Rightarrow \frac{4}{5}\text{tan}^2 60° - \frac{2}{\text{sin}^2 30°} - \frac{3}{4}\text{tan}^2 30° \\[1em] \Rightarrow \frac{4}{5} \times (\sqrt{3})^2 - \frac{2}{\Big(\frac{1}{2}\Big)^2} - \frac{3}{4} \times \Big(\frac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \frac{4}{5} \times 3 - 2 \times (2)^2 - \frac{3}{4} \times \frac{1}{3} \\[1em] \Rightarrow \frac{12}{5} - 8 - \frac{1}{4} \\[1em] \Rightarrow \frac{48 - 160 - 5}{20} \\[1em] \Rightarrow -\frac{117}{20}\\[1em] \Rightarrow -5\frac{17}{20}.\end{aligned}$$

Hence, $\frac{4}{5}\text{tan}^2 60° - \frac{2}{\text{sin}^2 30°} - \frac{3}{4}\text{tan}^2 30° = -5\frac{17}{20}$.

(i) Solving,

L.H.S. of the equation : cos² 30° + sin 30° + tan² 45° = $2\frac{1}{4}$.

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{3}}{2}\Big)^2 + \frac{1}{2} + (1)^2 \\[1em] \Rightarrow \frac{3}{4} + \frac{1}{2} + 1 \\[1em] \Rightarrow \frac{3 + 2 + 4}{4} \\[1em] \Rightarrow \frac{9}{4} \\[1em] \Rightarrow 2\frac{1}{4}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cos² 30° + sin 30° + tan² 45° = $2\frac{1}{4}$.

(ii) Solving,

L.H.S. of the equation : 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) = 2

$$\begin{aligned}\Rightarrow 4\Big[\Big(\frac{1}{2}\Big)^4 + \Big(\frac{1}{2}\Big)^4\Big] - 3\Big[\Big(\frac{1}{\sqrt{2}}\Big)^2 - (1)^2\Big] \\[1em] \Rightarrow 4\Big[\frac{1}{16} + \frac{1}{16}\Big] - 3\Big[\frac{1}{2} - 1\Big] \\[1em] \Rightarrow 4 \times \frac{2}{16} - 3 \times -\frac{1}{2} \\[1em] \Rightarrow \frac{1}{2} + \frac{3}{2} \\[1em] \Rightarrow \frac{4}{2} \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

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

(iii) Solving,

R.H.S. of the equation : cos 60° = cos² 30° - sin² 30°.

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

Since, L.H.S. = R.H.S.

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

To verify,

tan 2x = $\frac{\text{2 tan x}}{\text{1 - tan}^2 x}$.

Substituting value of x in L.H.S. of the above equation,

⇒ tan 2x = tan 2(30°) = tan 60° = $\sqrt{3}$.

Substituting value of x in R.H.S. of the equation.

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

Since, L.H.S. = R.H.S.

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

To verify,

4 sin 2x cos 4x sin 6x = 1

Substituting value of x in L.H.S. of the above equation.

⇒ 4 sin 2(15°) cos 4(15°) sin 6(15°)

⇒ 4 sin 30° cos 60° sin 90°

⇒ $4 \times \frac{1}{2} \times \frac{1}{2} \times 1$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that 4 sin 2x cos 4x sin 6x = 1.

(i) Solving,

$$\begin{aligned}\Rightarrow \sqrt{\frac{1 - \text{cos}^2 30°}{1 - \text{sin}^2 30°}} \\[1em] \Rightarrow \sqrt{\frac{1 - \Big(\frac{\sqrt{3}}{2}\Big)^2}{1 - \Big(\frac{1}{2}\Big)^2}} \\[1em] \Rightarrow \sqrt{\frac{1 - \frac{3}{4}}{1 - \frac{1}{4}}} \\[1em] \Rightarrow \sqrt{\frac{\frac{1}{4}}{\frac{3}{4}}} \\[1em] \Rightarrow \sqrt{\frac{1 \times 4}{3 \times 4}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}}.\end{aligned}$$

Hence, $\sqrt{\frac{1 - \text{cos}^2 30°}{1 - \text{sin}^2 30°}} = \frac{1}{\sqrt{3}}$.

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin 45° cos 45° cos 60°}}{\text{sin 60° cos 30° tan 45°}} \\[1em] \Rightarrow \frac{\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}}{\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \times 1} \\[1em] \Rightarrow \frac{\frac{1}{4}}{\frac{3}{4}} \\[1em] \Rightarrow \frac{1}{3}.\end{aligned}$$

Hence, $\frac{\text{sin 45° cos 45° cos 60°}}{\text{sin 60° cos 30° tan 45°}} = \frac{1}{3}.$

(i) To verify,

sin 2θ = 2 sin θ cos θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 2θ = sin 2(30°) = sin 60° = $\frac{\sqrt{3}}{2}$.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2 sin θ cos θ = 2 sin 30° cos 30° = $2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that sin 2θ = 2 sin θ cos θ.

(ii) To verify,

cos 2θ = 2 cos² θ - 1

Substituting value of θ in L.H.S. of the equation we get :

⇒ cos 2θ = cos 2(30°) = cos 60° = $\frac{1}{2}$.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 2cos² θ - 1 = 2cos² 30° - 1

= $2 \times \Big(\frac{\sqrt{3}}{2}\Big)^2 - 1$

= $2 \times \frac{3}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that cos 2θ = 2cos² θ - 1.

(iii) To verify,

sin 3θ = 3 sin θ - 4 sin³ θ

Substituting value of θ in L.H.S. of the equation we get :

⇒ sin 3θ = sin 3(30°) = sin 90° = 1.

Substituting value of θ in R.H.S. of the equation we get :

⇒ 3 sin θ - 4 sin³ θ = 3 sin 30° - 4 sin³ 30°

= $3 \times \frac{1}{2} - 4 \times \Big(\frac{1}{2}\Big)^3$

= $\frac{3}{2} - \frac{4}{8}$

= $\frac{12 - 4}{8} = \frac{8}{8}$ = 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 3θ = 3 sin θ - 4 sin³ θ.

(iv) Given,

Equation : cos 3θ = 4 cos³θ - 3 cos θ

Substituting θ = 30°, in L.H.S. of the given equation, we get :

⇒ cos 3θ = cos 3(30°) = cos 90° = 0.

Substituting θ = 30°, in R.H.S. of the given equation, we get :

⇒ 4 cos³θ - 3 cos θ = 4 cos³ 30° - 3 cos 30°

$$\begin{aligned}= 4 \times \Big(\frac{\sqrt{3}}{2}\Big)^3 - 3 \times \frac{\sqrt{3}}{2}\\[1em] = 4 \times \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} \\[1em] = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \\[1em] = 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cos 3θ = 4 cos³θ - 3 cos θ.

Substituting, θ = 30° in 2 sin θ : sin 2θ we get,

⇒ 2 sin 30° : sin 2(30°)

⇒ 2 sin 30° : sin 60°

⇒ $\frac{2 \times \frac{1}{2}}{\frac{\sqrt{3}}{2}}$

⇒ $\frac{1}{\frac{\sqrt{3}}{2}}$

⇒ $\frac{2}{\sqrt{3}} = 2 : \sqrt{3}$.

Hence, 2 sin θ : sin 2θ = $\frac{2}{\sqrt{3}} = 2 : \sqrt{3}$.

Let A = 30° and B = 60°.

Substituting values in sin(A + B) we get :

⇒ sin(A + B) = sin(30° + 60°) = sin 90° = 1.

Substituting values in sin A + sin B we get :

⇒ sin A + sin B = sin 30° + sin 60°

= $\frac{1}{2} + \frac{\sqrt{3}}{2}$

= $\frac{1 + \sqrt{3}}{2}$.

As, LHS ≠ RHS

Hence, proved that sin(A + B) ≠ sin A + sin B.

(i) To verify,

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

Substituting values in L.H.S. of the above equation :

sin(A + B) = sin(60° + 30°) = sin 90° = 1.

Substituting values in R.H.S. of the equation :

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

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

Since, L.H.S. = R.H.S.

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

(ii) To verify,

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

Substituting values in L.H.S. of equation :

cos(A + B) = cos(60° + 30°) = cos 90° = 0.

Substituting values in R.H.S. of equation :

cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

$$\begin{aligned}= \frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \\[1em] = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\[1em] = 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

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

(iii) To verify,

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

Substituting values in L.H.S. of equation :

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

Substituting values in R.H.S. of equation :

sin A cos B - cos A sin B = sin 60° cos 30° - cos 60° sin 30°

$$\begin{aligned}= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} \\[1em] = \frac{3}{4} - \frac{1}{4} \\[1em] = \frac{2}{4} = \frac{1}{2}.\end{aligned}$$

Since, L.H.S. = R.H.S.

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

(iv) To verify,

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

Substituting values in L.H.S. of equation :

tan (A - B) = tan (60° - 30°) = tan 30° = $\frac{1}{\sqrt{3}}$.

Substituting values in R.H.S. of equation :

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

Since, L.H.S. = R.H.S.

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

Given,

⇒ 2 sin 2θ = $\sqrt{3}$

⇒ sin 2θ = $\frac{\sqrt{3}}{2}$

⇒ sin 2θ = sin 60°

⇒ 2θ = 60°

⇒ θ = 30°.

Hence, θ = 30°.

We know that,

sin 60° = $\frac{\sqrt{3}}{2}$ = cos 30°.

Given,

⇒ cos(20° + x) = sin 60°

⇒ cos(20° + x) = cos 30°

⇒ (20° + x) = 30°

⇒ x = 10°.

Hence, x = 10°.

Given,

$$\begin{aligned}\Rightarrow 3\text{sin}^2 \text{ θ} = \frac{9}{4} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{3}{4} \\[1em] \Rightarrow \text{sin} \text{ θ} = \pm \frac{\sqrt{3}}{2}.\end{aligned}$$

Since, θ is an acute angle.

∴ sin θ = $\frac{\sqrt{3}}{2}.$

⇒ sin θ = sin 60°

⇒ θ = 60°.

Hence, θ = 60°.

Given,

sin θ = cos θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Substituting value in 2 tan² θ + sin² θ - 1 we get :

⇒ 2 tan² 45° + sin² 45° - 1

⇒ 2(1)² + $\Big(\frac{1}{\sqrt{2}}\Big)^2 -1$

⇒ 2 + $\frac{1}{2}$ - 1

⇒ $1\frac{1}{2}$.

Hence, 2 tan² θ + sin² θ - 1 = $1\frac{1}{2}$.

(i) By formula,

tan x° = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} = \sqrt{3}$.

Hence, tan x° = $\sqrt{3}$.

(ii) tan x° = $\sqrt{3}$

⇒ tan x° = tan 60°

⇒ x = 60.

Hence, x = 60.

(iii) Substituting value of x in cos x°, we get :

⇒ cos x° = cos 60°

⇒ cos x° = $\frac{1}{2}$.

Hence, cos x° = $\frac{1}{2}$.

(iv) Substituting value of x in sin x°, we get :

⇒ sin x° = sin 60° = $\frac{\sqrt{3}}{2}$.

By formula,

$$\begin{aligned}\Rightarrow \text{sin x}^{\circ} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{AB}{AC} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{y} \\[1em] \Rightarrow y = 2.\end{aligned}$$

Hence, y = 2.

(i) Given,

⇒ 2 sin 3θ = $\sqrt{3}$

⇒ sin 3θ = $\frac{\sqrt{3}}{2}$

⇒ sin 3θ = sin 60°

⇒ 3θ = 60°

⇒ θ = $\frac{60^{\circ}}{3}$

⇒ θ = 20°.

Hence, θ = 20°.

(ii) Given,

⇒ tan 3θ = 1

⇒ tan 3θ = tan 45°

⇒ 3θ = 45°

⇒ θ = $\frac{45^{\circ}}{3}$

⇒ θ = 15°.

Hence, θ = 15°.

Given,

$$\begin{aligned}\Rightarrow \text{tan 3x = sin 45° cos 45° + sin 30°} \\[1em] \Rightarrow \text{tan 3x} = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \\[1em] \Rightarrow \text{tan 3x} = \frac{1}{2} + \frac{1}{2} \\[1em] \Rightarrow \text{tan 3x} = 1 \\[1em] \Rightarrow \text{tan 3x = tan 45°} \\[1em] \Rightarrow 3x = 45° \\[1em] \Rightarrow x = \frac{45^{\circ}}{3} \\[1em] \Rightarrow x = 15°.\end{aligned}$$

Hence, x = 15°.

(i) Given,

$$\begin{aligned}\Rightarrow \text{4 cos}^2 x° - 1 = 0 \\[1em] \Rightarrow \text{4 cos}^2 x° = 1 \\[1em] \Rightarrow \text{cos}^2 x° = \frac{1}{4} \\[1em] \Rightarrow \text{cos x°} = \sqrt{\frac{1}{4}} \\[1em] \Rightarrow \text{cos x°} = \pm \frac{1}{2}\end{aligned}$$

Since, x is an acute angle.

$\therefore \text{cos x°} = \frac{1}{2}$.

⇒ cos x° = cos 60°

⇒ x = 60.

Hence, x = 60.

(ii) Substituting value of x in sin² x° + cos² x° we get :

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

Hence, sin² x° + cos² x° = 1.

(iii) Substituting value of x in cos² x° - sin² x° we get :

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

Hence, cos² x° - sin² x° = $-\frac{1}{2}$.

Given,

$$\begin{aligned}\Rightarrow \text{sec θ = cosec θ} \\[1em] \Rightarrow \frac{1}{\text{cos θ}} = \frac{1}{\text{sin θ}} \\[1em] \Rightarrow \frac{\text{sin θ}}{\text{cos θ}} = 1 \\[1em] \Rightarrow \text{tan θ} = 1 \\[1em] \Rightarrow \text{tan θ = tan 45°} \\[1em] \Rightarrow \text{θ} = 45°.\end{aligned}$$

Hence, θ = 45°.

Given,

As, θ is in the range of 0° ≤ θ ≤ 90°.

tan θ = cot θ [Only when θ = 45°].

Hence, θ = 45°.

Given,

⇒ sin 3x = 1

⇒ sin 3x = sin 90°

⇒ 3x = 90°

⇒ x = $\frac{90}{3}$

⇒ x = 30°.

(i) Substituting value of x in sin x, we get :

⇒ sin x = sin 30° = $\frac{1}{2}$.

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

(ii) Substituting value of x in cos 2x, we get :

⇒ cos 2x = cos 60° = $\frac{1}{2}$.

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

(iii) Substituting value of x in tan² x - sec² x, we get :

$$\begin{aligned}\Rightarrow \text{tan}^2 x - \text{sec}^2 x = \text{tan}^2 60° - \text{sec}^2 60° \\[1em] = (\sqrt{3})^2 - (2)^2 \\[1em] = 3 - 4 \\[1em] = -1.\end{aligned}$$

Hence, tan² x - sec² x = -1.

Given,

⇒ 3 tan² θ - 1 = 0

⇒ tan² θ = $\frac{1}{3}$

⇒ tan θ = $\sqrt{\frac{1}{3}}$

⇒ tan θ = $\pm \frac{1}{\sqrt{3}}$

As, θ is acute.

∴ tan θ = $\frac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

cos 2θ = cos 60° = $\frac{1}{2}$

Hence, cos 2θ = $\frac{1}{2}$.

Given,

⇒ sin x + cos y = 1

⇒ sin 30° + cos y = 1

⇒ $\frac{1}{2}$ + cos y = 1

⇒ cos y = 1 - $\frac{1}{2}$

⇒ cos y = $\frac{1}{2}$

⇒ cos y = cos 60°.

Hence, y = 60°.

Given,

⇒ sin(A + B) = $\frac{\sqrt{3}}{2}$

⇒ sin(A + B) = sin 60°

⇒ A + B = 60° ..........(1)

Also,

⇒ cos(A - B) = $\frac{\sqrt{3}}{2}$

⇒ cos(A - B) = cos 30°

⇒ A - B = 30° ..........(2)

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

⇒ (A + B) + (A - B) = 60° + 30°

⇒ A + A + B - B = 90°

⇒ 2A = 90°

⇒ A = 45°.

Substituting value of A in (1), we get :

⇒ A + B = 60°

⇒ 45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°.

Hence, A = 45° and B = 15°.

We know that the diagonals of a rhombus bisect the opposite angles and are perpendicular to each other.

∴ ∠OAB = $\frac{60°}{2}$ = 30°.

In right ∠AOB,

⇒ sin 30° = $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{OB}{AB}$

⇒ $\frac{1}{2} = \frac{OB}{AB}$

⇒ OB = $\frac{AB}{2}$

⇒ OB = $\frac{8}{2}$ = 4 cm.

As diagonals of rhombus bisect each other.

∴ BD = 2 OB = 2 × 4 = 8 cm.

cos 30° = $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{OA}{AB}$

⇒ $\frac{\sqrt{3}}{2} = \frac{OA}{AB}$

⇒ OA = $\frac{AB\sqrt{3}}{2}$

⇒ OA = $\frac{8\sqrt{3}}{2} = 4\sqrt{3}$.

As diagonals of rhombus bisect each other.

∴ AC = 2 OA = $2 \times 4\sqrt{3} = 8\sqrt{3}$.

Hence, the length of the diagonals of the rhombus are 8 cm and $8\sqrt{3}$ cm.

From figure,

$$\begin{aligned}\text{sin 60°} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{AC}{AB} \\[1em] \Rightarrow \frac{\sqrt{3}}{2} = \frac{6}{AB} \\[1em] \Rightarrow AB = \frac{12}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AB = \frac{12\sqrt{3}}{3} \\[1em] \Rightarrow AB = 4\sqrt{3} \text{ cm}. \\[1em] \text{tan 60°} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \frac{AC}{BC} \\[1em] \Rightarrow \sqrt{3} = \frac{6}{BC} \\[1em] \Rightarrow BC = \frac{6}{\sqrt{3}} = 2\sqrt{3}.\end{aligned}$$

Hence, AB = $4\sqrt{3}$ cm and BC = $2\sqrt{3}$ cm.

Let AB = d meters, then PC = d meters.

From right-angled △PCQ,

tan 30° = $\frac{CQ}{PC}$

⇒ $\frac{1}{\sqrt{3}} = \frac{BQ - BC}{d}$

From figure,

BC = AP = 1.8 m

⇒ $\frac{1}{\sqrt{3}} = \frac{13.8 - 1.8}{d}$

⇒ $\frac{1}{\sqrt{3}} = \frac{12}{d}$

⇒ d = $12\sqrt{3}$ meters.

Hence, distance of man from building is $12\sqrt{3}$ meters.

In △ABD,

⇒ tan 45° = $\frac{AD}{BD}$

⇒ 1 = $\frac{AD}{BD}$

⇒ BD = AD.

In △ADC,

⇒ tan 60° = $\frac{AD}{DC}$

⇒ $\sqrt{3} = \frac{AD}{DC}$

⇒ DC = $\frac{AD}{\sqrt{3}}$

From figure,

BC = BD + DC

$$\begin{aligned}\Rightarrow 8 = AD + \frac{AD}{\sqrt{3}} \\[1em] \Rightarrow 8 = \frac{\sqrt{3}AD + AD}{\sqrt{3}} \\[1em] \Rightarrow \frac{AD(\sqrt{3} + 1)}{\sqrt{3}} = 8 \\[1em] \Rightarrow AD = \frac{8\sqrt{3}}{\sqrt{3} + 1}\end{aligned}$$

Multiplying numerator and denominator by $(\sqrt{3} - 1)$

$$\begin{aligned}\Rightarrow AD = \frac{8\sqrt{3}}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow AD = \frac{8(3 - \sqrt{3})}{(\sqrt{3})^2 - (1)^2} \space [\because a^2 - b^2 = (a+b)(a-b)] \\[1em] \Rightarrow AD = \frac{8(3 - \sqrt{3})}{3 - 1} \\[1em] \Rightarrow AD = \frac{8(3 - \sqrt{3})}{2}\\[1em] \Rightarrow AD = 4(3 - \sqrt{3}) \text{ m}\end{aligned}$$

Hence, AD = $4(3 - \sqrt{3})$ m.

(i) Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cos 18°}}{\text{sin 72°}} \\[1em] \Rightarrow \frac{\text{cos 18°}}{\text{sin (90° - 18°)}} \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \frac{\text{cos 18°}}{\text{cos 18°}} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{cos 18°}}{\text{sin 72°}}$ = 1.

(ii) Solving,

$$\begin{aligned}\Rightarrow \frac{\text{tan 41°}}{\text{cot 49°}} \\[1em] \Rightarrow \frac{\text{tan (90° - 49°)}}{\text{cot 49°}} \\[1em] \text{As, tan (90 - θ) = cot θ} \\[1em] \Rightarrow \frac{\text{cot 49°}}{\text{cot 49°}} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{tan 41°}}{\text{cot 49°}} = 1$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow \frac{\text{cosec (90° - 72° 30}')}{\text{sec 72° 30}'} \\[1em] \text{As, cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \frac{\text{sec 72° 30}'}{\text{sec 72° 30}'} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{cosec 17° 30}'}{\text{sec 72° 30}'} = 1$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cot 40°}}{\text{tan 50°}} - \frac{1}{2}\Big(\frac{\text{cos 35°}}{\text{sin 55°}}\Big) \\[1em] \Rightarrow \frac{\text{cot 40°}}{\text{tan (90° - 40°)}} - \frac{1}{2}\Big(\frac{\text{cos 35°}}{\text{sin (90° - 35°)}}\Big) \\[1em] \text{As, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \frac{\text{cot 40°}}{\text{cot 40°}} - \frac{1}{2}\Big(\frac{\text{cos 35°}}{\text{cos 35°}}\Big) \\[1em] \Rightarrow 1 - \frac{1}{2} \\[1em] \Rightarrow \frac{1}{2}.\end{aligned}$$

Hence, $\frac{\text{cot 40°}}{\text{tan 50°}} - \frac{1}{2}\Big(\frac{\text{cos 35°}}{\text{sin 55°}}\Big) = \frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\frac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{\text{sin (90° - 41°)}}{\text{cos 41°}}\Big)^2 + \Big(\frac{\text{cos 41°}}{\text{sin (90° - 41°)}}\Big)^2 \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \Big(\frac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 + \Big(\frac{\text{cos 41°}}{\text{cos 41°}}\Big)^2 \\[1em] \Rightarrow 1^2 + 1^2 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, $\Big(\frac{\text{sin 49°}}{\text{cos 41°}}\Big)^2 + \Big(\frac{\text{cos 41°}}{\text{sin 49°}}\Big)^2 = 2.$

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin 72°}}{\text{cos 18°}} - \frac{\text{sec 32°}}{\text{cosec 58°}} \\[1em] \Rightarrow \frac{\text{sin (90° - 18°)}}{\text{cos 18°}} - \frac{\text{sec 32°}}{\text{cosec (90° - 32°)}} \\[1em] \text{As, cosec (90 - θ) = sec θ and sin (90 - θ) = cos θ} \\[1em] \Rightarrow \frac{\text{cos 18°}}{\text{cos 18°}} - \frac{\text{sec 32°}}{\text{sec 32°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $\frac{\text{sin 72°}}{\text{cos 18°}} - \frac{\text{sec 32°}}{\text{cosec 58°}} = 0.$

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cos 75°}}{\text{sin 15°}} + \frac{\text{sin 12°}}{\text{cos 78°}} - \frac{\text{cos 18°}}{\text{sin 72°}} \\[1em] \Rightarrow \frac{\text{cos 75°}}{\text{sin (90° - 75°)}} + \frac{\text{sin (90° - 78°)}}{\text{cos 78°}} - \frac{\text{cos 18°}}{\text{sin (90° - 18°)}} \\[1em] \text{As, sin (90 - θ) = cos θ} \\[1em] \Rightarrow \frac{\text{cos 75°}}{\text{cos 75°}} + \frac{\text{cos 78°}}{\text{cos 78°}} - \frac{\text{cos 18°}}{\text{cos 18°}} \\[1em] \Rightarrow 1 + 1 - 1 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{cos 75°}}{\text{sin 15°}} + \frac{\text{sin 12°}}{\text{cos 78°}} - \frac{\text{cos 18°}}{\text{sin 72°}} = 1.$

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin 25°}}{\text{sec 65°}} + \frac{\text{cos 25°}}{\text{cosec 65°}} \\[1em] \Rightarrow \frac{\text{sin 25°}}{\frac{1}{\text{cos 65°}}} + \frac{\text{cos 25°}}{\frac{1}{\text{sin 65°}}} \\[1em] \Rightarrow \text{sin 25°.cos 65° + cos 25°.sin 65°} \\[1em] \Rightarrow \text{sin 25°. cos(90° - 25°) + cos 25°. sin(90° - 25°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \text{sin 25°. sin 25° + cos 25°. cos 25°} \\[1em] \Rightarrow \text{sin}^2 25° + \text{cos}^2 25° \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{sin 25°}}{\text{sec 65°}} + \frac{\text{cos 25°}}{\text{cosec 65°}} = 1$.

Solving,

⇒ sin 62° - cos 28°

⇒ sin (90° - 28°) - cos 28°

⇒ cos 28° - cos 28° [As, sin (90 - θ) = cos θ]

⇒ 0.

Hence, sin 62° - cos 28° = 0.

Solving,

⇒ cosec 35° - sec 55°

⇒ cosec (90° - 55°) - sec 55°

⇒ sec 55° - sec 55° [As, cosec (90 - θ) = sec θ]

⇒ 0.

Hence, cosec 35° - sec 55° = 0.

Solving,

$$\begin{aligned}\Rightarrow \text{cos}^2 26° + \text{cos 64° sin 26°} + \frac{\text{tan 36°}}{\text{cot 54°}} \\[1em] \Rightarrow \text{cos}^2 26° + \text{cos (90° - 26°) sin 26°} + \frac{\text{tan (90° - 54°)}}{\text{cot 54°}} \\[1em] \text{As, cos (90 - θ) = sin θ and tan (90 - θ) = cot θ} \\[1em] \Rightarrow \text{cos}^2 26° + \text{sin}^2 26° + \frac{\text{cot 54°}}{\text{cot 54°}} \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, cos² 26° + cos 64° sin 26° + $\frac{\text{tan 36°}}{\text{cot 54°}}$ = 2.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sec 17°}}{\text{cosec (90° - 17°)}} + \frac{\text{tan (90° - 22°)}}{\text{cot 22°}} + \text{cos}^2 44° + \text{cos}^2 (90° - 44°) \\[1em] \Rightarrow \frac{\text{sec 17°}}{\text{sec 17°}} + \frac{\text{cot 22°}}{\text{cot 22°}} + \text{cos}^2 44° + \text{sin}^2 44° \\[1em] \Rightarrow 1 + 1 + 1 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $\frac{\text{sec 17°}}{\text{\text{cosec 73°}}} + \frac{\text{tan 68°}}{\text{cot 22°}}$ + cos² 44° + cos² 46° = 3.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cos 65°}}{\text{sin (90° - 65°)}} + \frac{\text{cos 32°}}{\text{sin (90° - 32°)}} - \text{sin 28°} \times \frac{1}{\text{cos 62°}} + \frac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow \frac{\text{cos 65°}}{\text{cos 65°}} + \frac{\text{cos 32°}}{\text{cos 32°}} - \text{sin 28°} \times \frac{1}{\text{cos 62°}} + \frac{1}{\text{sin}^2 30°} \\[1em] \Rightarrow 1 + 1 - \text{sin 28°} \times \frac{1}{cos(90° - 28°)} + \frac{1}{\Big(\frac{1}{2}\Big)^2} \\[1em] \Rightarrow 2 - \frac{\text{sin 28°}}{\text{sin 28°}} + 4 \\[1em] \Rightarrow 2 - 1 + 4 \\[1em] \Rightarrow 5.\end{aligned}$$

Hence, $\frac{\text{cos 65°}}{\text{sin 25°}} + \frac{\text{cos 32°}}{\text{sin 58°}} - \text{sin 28° sec 62° + cosec}^2 30°$ = 5.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°) \\[1em] \Rightarrow \frac{\text{sec 29°}}{\text{cosec (90° - 29°)}} + \text{2 cot (90° - 82°) cot (90° - 73°) cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 (90° - 38°)) \\[1em] \text{As, cosec(90° - θ) = sec θ, cot(90° - θ) = tan θ} \\[1em] \Rightarrow \frac{\text{sec 29°}}{\text{sec 29°}} + \text{2 tan 82° tan 73° cot 45°} \times \frac{1}{\text{tan 73°}} \times \frac{1}{\text{tan 82°}} - \text{3 (sin}^2 38° + \text{cos}^2 38°) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \\[1em] \Rightarrow 1 + \text{2 cot 45°} - 3 \\[1em] \Rightarrow 1 + 2 - 3 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $\frac{\text{sec 29°}}{\text{cosec 61°}} + \text{2 cot 8° cot 17° cot 45° cot 73° cot 82°} - \text{3 (sin}^2 38° + \text{sin}^2 52°)$ = 0.

Solving,

⇒ tan 81° + cos 72°

⇒ tan (90° - 9°) + cos (90° - 18°)

As, tan(90° - θ) = cot θ, cos(90° - θ) = sin θ

⇒ cot 9° + sin 18°.

Hence, tan 81° + cos 72° = cot 9° + sin 18°.

Solving,

⇒ cot 49° + cosec 87°

⇒ cot (90° - 41°) + cosec (90° - 3°)

As, cot(90° - θ) = tan θ, cosec(90° - θ) = sec θ

⇒ tan 41° + sec 3°.

Hence, cot 49° + cosec 87° = tan 41° + sec 3°.

To prove,

sin² 28° - cos² 62° = 0.

Solving, L.H.S. of the equation.

sin² 28° - cos² 62°

= sin² 28° - cos² (90° - 28°)

= sin² 28° - sin² 28°

= 0.

Since, L.H.S. = R.H.S.

Hence, proved that sin² 28° - cos² 62° = 0.

To prove,

cos² 25° + cos² 65° = 1.

Solving, L.H.S. of the equation.

cos² 25° + cos² 65°

= cos² 25° + cos² (90° - 25°)

As, cos (90° - θ) = sin θ

= cos² 25° + sin² 25°

= 1 [∵ cos² θ + sin² θ = 1]

Since, L.H.S. = R.H.S.

Hence, proved that cos² 25° + cos² 65° = 1.

To prove,

cosec² 67° - tan² 23° = 1

Solving L.H.S. of the equation,

$$\begin{aligned}\phantom{\Rightarrow} \text{cosec}^2 67° - \text{tan}^2 23° \\[1em] \Rightarrow \frac{1}{\text{sin}^2 67°} - \text{tan}^2 (90° - 67°) \\[1em] \Rightarrow \frac{1}{\text{sin}^2 67°} - \text{cot}^2 67° \\[1em] \Rightarrow \frac{1}{\text{sin}^2 67°} - \frac{\text{cos}^2 67°}{\text{sin}^2 67°} \\[1em] \Rightarrow \frac{\text{1 - cos}^2 67°}{\text{sin}^2 67°} \\[1em] \Rightarrow \frac{\text{sin}^2 67°}{\text{sin}^2 67°} \space [\because \text{ 1 - cos}^2 \text{ θ} = \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cosec² 67° - tan² 23° = 1.

To prove,

sec² 22° - cot² 68° = 1

Solving, L.H.S. of the equation we get :

sec² 22° - cot² 68°

⇒ sec² 22° - cot² (90° - 22°)

We know that,

cot (90 - θ) = tan θ

⇒ sec² 22° - tan² 22°

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos}^2 22°} - \frac{\text{sin}^2 22°}{\text{cos}^2 22°} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 22°}{\text{cos}^2 22°} \\[1em] \text{As, 1 - sin}^2 θ = \text{cos}^2 θ \\[1em] \Rightarrow \frac{\text{cos}^2 22°}{\text{cos}^2 22°} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sec² 22° - cot² 68° = 1.

To prove,

sin 63° cos 27° + cos 63° sin 27° = 1

Solving, L.H.S. of the equation we get,

⇒ sin (90° - 27°) cos 27° + cos (90° - 27°) sin 27°

We know that,

sin (90 - θ) = cos θ and cos (90 - θ) = sin θ

⇒ cos 27° cos 27° + sin 27° sin 27°

⇒ cos² 27° + sin² 27°

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 63° cos 27° + cos 63° sin 27° = 1.

To prove,

sec 31° sin 59° + cos 31° cosec 59° = 2

Solving, L.H.S. of the equation we get,

⇒ sec 31° sin 59° + cos 31° cosec 59

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos 31°}} \times \text{sin (90° - 31°)} + \text{cos (90° - 59°)} \times \frac{1}{\text{sin 59°}} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \frac{1}{\text{cos 31°}} \times \text{cos 31°} + \text{sin 59°} \times \frac{1}{\text{sin 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sec 31° sin 59° + cos 31° cosec 59° = 2.

To prove,

sec 70° sin 20° - cos 20° cosec 70° = 0

Solving L.H.S. of the equation we get :

$$\begin{aligned}\Rightarrow \text{sec 70° sin 20° - cos 20° cosec 70°} \\[1em] \Rightarrow \frac{1}{\text{cos 70°}} \times \text{sin (90° - 70°)} - \text{cos (90° - 70°)} \times \frac{1}{\text{sin 70°}} \\[1em] \text{As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ} \\[1em] \Rightarrow \frac{1}{\text{cos 70°}} \times \text{cos 70°} - \text{sin 70°} \times \frac{1}{\text{sin 70°}} \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sec 70° sin 20° - cos 20° cosec 70° = 0.

To prove,

sin² 20° + sin² 70° - tan² 45° = 0.

Solving, L.H.S. of the equation we get :

sin² 20° + sin² 70° - tan² 45°

We know that,

sin (90 - θ) = cos θ

and

sin² θ + cos² θ = 1.

⇒ sin² 20° + sin² (90° - 20°) - (1)²

⇒ sin² 20° + cos² 20° - 1

⇒ 1 - 1

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that sin² 20° + sin² 70° - tan² 45° = 0.

To prove,

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

Solving, L.H.S. of the equation we get :

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

Since, L.H.S. = R.H.S.

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

To prove,

$\frac{\text{sin 50°}}{\text{cos 40°}} + \frac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0$.

Solving, L.H.S. of the equation we get :

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{sin 50°}}{\text{cos 40°}} + \frac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0. \\[1em] \Rightarrow \frac{\text{sin (90° - 40°)}}{\text{cos 40°}} + \frac{\text{cosec (90° - 50°)}}{\text{sec 50°}} - \text{4 cos 50° cosec (90° - 50°) + 2} = 0 \\[1em]\end{aligned}$$

We know that,

sin (90 - θ) = cos θ

and

cosec (90 - θ) = sec θ

$$\begin{aligned}\Rightarrow \frac{\text{cos 40°}}{\text{cos 40°}} + \frac{\text{sec 50°}}{\text{sec 50°}} - \text{4 cos 50° sec 50°} + 2 \\[1em] \Rightarrow 1 + 1 - 4 \times \text{cos 50°} \times \frac{1}{\text{cos 50°}} + 2 \\[1em] \Rightarrow 2 - 4 + 2 \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin 50°}}{\text{cos 40°}} + \frac{\text{cosec 40°}}{\text{sec 50°}} - \text{4 cos 50° cosec 40° + 2} = 0$.

To prove,

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

Solving, L.H.S. of the equation we get :

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

Since, L.H.S. = R.H.S.

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

To prove,

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

Solving, L.H.S. of the equation we get :

$$\begin{aligned}\phantom{\Rightarrow} \frac{\text{cos 80°}}{\text{sin 10°}} + \text{cos 59° cosec 31°} \\[1em] \Rightarrow \frac{\text{cos 80°}}{\text{sin (90° - 80°)}} + \text{cos 59° cosec (90° - 59°)} \\[1em] \text{As, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ } \\[1em] \Rightarrow \frac{\text{cos 80°}}{\text{cos 80°}} + \text{cos 59° sec 59°} \\[1em] \Rightarrow 1 + \text{cos 59°} \times \frac{1}{\text{cos 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

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

Solving,

$$\begin{aligned}2\Big(\frac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\frac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\frac{\text{sec 40°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2\Big(\frac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\frac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big) - 3\Big(\frac{\text{sec (90° - 50°)}}{\text{cosec 50°}}\Big)\\[1em] \text{As, cot (90 - θ) = tan θ and sec (90 - θ) = cosec θ} \\[1em] \Rightarrow 2\Big(\frac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\frac{\text{tan 35°}}{\text{tan 35°}}\Big) - 3\Big(\frac{\text{cosec 50°}}{\text{cosec 50°}}\Big)\\[1em] \Rightarrow 2(1)^2 + 1 - 3 \\[1em] \Rightarrow 3 - 3 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, $2\Big(\frac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\frac{\text{cot 55°}}{\text{tan 35°}}\Big) - 3\Big(\frac{\text{sec 40°}}{\text{cosec 50°}}\Big) = 0$

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\frac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 \space (90° - 80°) - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin (90 - θ) = cos θ, cos (90 - θ) = sin θ and cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \Big(\frac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\frac{\text{sin}^2 \space 35° + \text{cos}^2 \space 35°}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \text{ and } \text{sec}^2 θ - \text{tan}^2 θ = 1 \\[1em] \Rightarrow \frac{1}{1} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\Big(\frac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big)$ = 1.

Solving,

⇒ sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

⇒ sin² 34° + sin² (90° - 34°) + 2 tan (90° - 72°) tan 72° - cot² 30°

We know that,

sin (90° - θ) = cos θ, cos (90° - θ) = sin θ and tan (90° - θ) = cot θ

⇒ sin² 34° + cos² 34° + 2 cot 72° tan 72° - cot² 30°

As,

cot θ. tan θ = 1 and sin² θ + cos² θ = 1.

⇒ 1 + 2 - $(\sqrt{3})^2$

⇒ 3 - 3

⇒ 0.

Hence, sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30 = 0.

To prove,

$\frac{\text{cos θ}}{\text{sin (90° - θ)}} + \frac{\text{sin θ}}{\text{cos (90° - θ)}} = 2$

We know that,

sin (90 - θ) = cos θ and cos (90 - θ) = sin θ

Solving L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{cos θ}}{\text{cos θ}} + \frac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos θ}}{\text{sin (90° - θ)}} + \frac{\text{sin θ}}{\text{cos (90° - θ)}} = 2$.

To prove,

cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.

We know that,

sin (90 - θ) = cos θ and cos (90 - θ) = sin θ

Solving L.H.S. of the equation, we get :

⇒ cos θ cos θ + sin θ sin θ

⇒ cos² θ + sin² θ

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.

To prove,

$\frac{\text{tan θ}}{\text{\text{tan (90° - θ)}}} + \frac{\text{sin (90° - θ)}}{\text{cos θ}} = \text{sec}^2 \text{ θ}.$

We know that,

sin (90 - θ) = cos θ and tan (90 - θ) = cot θ

Solving L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{tan θ}}{\text{\text{cot θ}}} + \frac{\text{cos θ}}{\text{cos θ}} \\[1em] \Rightarrow \frac{\text{tan θ}}{\frac{1}{\text{tan θ}}} + 1 \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + 1 \\[1em] \Rightarrow \text{sec}^2 \text{ θ}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{tan θ}}{\text{\text{tan (90° - θ)}}} + \frac{\text{sin (90° - θ)}}{\text{cos θ}} = \text{sec}^2 \text{ θ}.$

We know that,

cos (90 - θ) = sin θ, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ.

Solving L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{sin A cos A}}{\text{cot A}} \\[1em] \Rightarrow \frac{\text{sin A cos A}}{\frac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \text{sin}^2 \text{ A} \\[1em] \Rightarrow 1 - \text{cos}^2 \text{ A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos (90° - A) sin (90° - A)}}{\text{tan} (90° - \text{ A})} = 1 - \text{cos}^ 2 \text{ A}$.

We know that,

cos (90 - θ) = sin θ, sec (90 - θ) = cosec θ, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ.

Solving L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{\text{sec A}} + \frac{\text{sin A}}{\text{cosec A}} \\[1em] \Rightarrow \frac{\text{cos A}}{\frac{1}{\text{cos A}}} + \frac{\text{sin A}}{\frac{1}{\text{sin A}}} \\[1em] \Rightarrow \text{cos}^2 A + \text{sin}^2 A \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin (90° - A)}}{\text{cosec (90° - A)}} + \frac{\text{cos (90° - A)}}{\text{sec (90° - A)}}$ = 1.

We know that,

sin (90° - θ) = cos θ, cos (90° - θ) = sin θ and sec (90° - θ) = cosec θ.

Substituting values in equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{cos θ}}{\text{sin (90° - θ)}} + \frac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30° \\[1em] \Rightarrow \frac{\text{cos θ}}{\text{cos θ}} + \frac{\text{\text{sin θ}}}{\text{cosec θ}} - \text{3 tan}^2 30° \\[1em] \Rightarrow 1 + \frac{\text{sin θ}}{\frac{1}{\text{sin θ}}} -3 \times \Big(\frac{1}{\sqrt{3}}\Big)^2 \\[1em] = 1 + \text{sin}^2 θ - 3 \times \frac{1}{3} \\[1em] = 1 - 1 + \text{sin}^2 θ\\[1em] = \text{sin}^2 θ.\end{aligned}$$

Hence, $\frac{\text{cos θ}}{\text{sin (90° - θ)}} + \frac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30°$ = sin² θ.

We know that,

sin (90° - θ) = cos θ
cos (90° - θ) = sin θ
sec (90° - θ) = cosec θ
cosec (90° - θ) = cot θ
cot (90° - θ) = tan θ.

Substituting values in equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \frac{\text{cot θ}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \frac{\text{sec θ cos θ tan θ}}{\text{sin θ cosec θ tan θ}} + \frac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow \frac{\frac{1}{\text{cos θ}} \times \text{cos θ} \times \text{tan θ}}{\text{sin θ} \times \frac{1}{\text{sin θ}} \times \text{tan θ}} + 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, $\frac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \frac{\text{cot θ}}{\text{tan (90° - θ)}}$ = 2.

Solving, L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} \\[1em] \Rightarrow \frac{\text{cos}^2 (45° + θ) + \text{sin}^2 [90° - (45° - θ)]}{\text{tan (60° + θ) \text{cot [90° - (30° - θ)]}}} \\[1em] \Rightarrow \frac{\text{cos}^2 (45° + θ) + \text{sin}^2 (45° + θ)}{\text{tan (60° + θ) \text{cot (60° + θ)}}} \\[1em] \text{As, cos}^2 \text{ A + sin}^2 A = 1 \text{ and tan A. cot A} = 1 \\[1em] \Rightarrow \frac{1}{1} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} = 1$.

Given,

sin 3A = cos (A - 6°)

⇒ sin 3A = sin [90° - (A - 6°)]

⇒ 3A = 90° - (A - 6°)

⇒ 3A = 96° - A

⇒ 4A = 96°

⇒ A = $\frac{96°}{4}$

⇒ A = 24°.

Hence, A = 24°.

Given,

tan 2A = cot (A - 18°)

⇒ tan 2A = tan [90° - (A - 18°)]

⇒ 2A = 90° - (A - 18°)

⇒ 2A = 90° - A + 18°

⇒ 2A = 108° - A

⇒ 3A = 108°

⇒ 3A = $\frac{108°}{3}$

⇒ A = 36°

Hence, A = 36°.

Given,

sec 2A = cosec (A - 27°)

⇒ sec 2A = sec [90° - (A - 27°)]

⇒ 2A = 90° - (A - 27°)

⇒ 2A = 90° - A + 27°

⇒ 2A = 117° - A

⇒ 3A = 117°

⇒ A = $\frac{117°}{3}$

⇒ A = 39°.

Hence, A = 39°.

Given,

cos 63° sec (90° - θ) = 1

We know that,

cos A sec A = 1

∴ 90° - θ = 63°

⇒ θ = 90° - 63°

⇒ θ = 27°.

Hence, θ = 27°.

Given,

tan 35° cot (90° - θ) = 1

We know that,

tan A cot A = 1

∴ 90° - θ = 35°

⇒ θ = 90° - 35°

⇒ θ = 55°.

Hence, θ = 55°.

(i) Given,

A, B and C are the interior angles of a △ABC.

∴ A + B + C = 180°

⇒ $\frac{A + B + C}{2} = 90°$

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

To prove,

cos $\frac{A + B}{2}$ = sin $\frac{C}{2}$

Substituting value of $\frac{A + B}{2}$ in L.H.S. of the equation we get :

$$\begin{aligned}\phantom{=} \text{cos } \Big(\frac{A + B}{2}\Big) \\[1em] = \text{cos } \Big(90^{\circ} - \frac{C}{2}\Big) \\[1em] = \text{sin } \frac{C}{2} \space [\because \text{cos (90 - θ)} = \text{sin θ}]\end{aligned}$$

Since, L.H.S. = R.H.S.

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

(ii) Given,

A, B and C are the interior angles of a △ABC.

∴ A + B + C = 180°

⇒ $\frac{A + B + C}{2} = 90°$

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

To prove,

tan $\frac{C + A}{2}$ = cot $\frac{B}{2}$

Substituting value of $\frac{C + A}{2}$ in L.H.S. of equation we get :

$$\begin{aligned}\phantom{=} \text{tan } \Big(\frac{C + A}{2}\Big) \\[1em] = \text{tan } \Big(90^{\circ} - \frac{B}{2}\Big) \\[1em] = \text{cot } \frac{B}{2} \space [\because \text{tan (90 - θ)} = \text{cot θ}]\end{aligned}$$

Since, L.H.S. = R.H.S.

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

Solving,

$$\begin{aligned}\Rightarrow \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, Option 4 is the correct option.

Solving,

$$\begin{aligned}\text{sin 45° + cos 45°} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \\[1em] = \frac{2}{\sqrt{2}} \\[1em] = \sqrt{2}.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \text{tan}^2 30° - 4\text{sin}^2 45° = \Big(\frac{1}{\sqrt{3}}\Big)^2 - 4 \times \Big(\frac{1}{\sqrt{2}}\Big)^2 \\[1em] = \frac{1}{3} - 4 \times \frac{1}{2} \\[1em] = \frac{1}{3} - 2 \\[1em] = \frac{1 - 6}{3} \\[1em] = -\frac{5}{3}.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \text{2 sin A cos A} = \text{2 sin 30° cos 30°} \\[1em] = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \\[1em] = \frac{\sqrt{3}}{{2}}.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\text{(sin 30° + cos 30°) - (sin 60° + cos 60°)} = \Big(\frac{1}{2} + \frac{\sqrt{3}}{{2}}\Big) - \Big(\frac{\sqrt{3}}{2} + \frac{1}{2}\Big) \\[1em] = \frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} - \frac{1}{2} \\[1em] = 0.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{3}\text{ cosec 60° - sec 60°} = \sqrt{3} \times \frac{2}{\sqrt{3}} - 2 \\[1em] = 2 - 2 \\[1em] = 0.\end{aligned}$$

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin 30°}} - \frac{\sqrt{3}}{\text{cos 30°}} = \frac{1}{\frac{1}{2}} - \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} \\[1em] = 2 - 2 \\[1em] = 0.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

⇒ tan A = $\sqrt{3}$

⇒ tan A = tan 60°

⇒ A = 60°.

⇒ cosec A = cosec 60° = $\frac{2}{\sqrt{3}}$.

Hence, Option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \text{sec θ. sin θ = 0} \\[1em] \Rightarrow \frac{1}{\text{cos θ}} \times \text{sin θ} = 0 \\[1em] \Rightarrow \text{tan θ} = 0 \\[1em] \Rightarrow θ = 0°.\end{aligned}$$

⇒ cos θ = cos 0° = 1.

Hence, Option 4 is the correct option.

Given,

⇒ sin α = $\frac{1}{2}$

⇒ sin α = sin 30°

⇒ α = 30°.

$$\begin{aligned}\text{3 cos α - 4 cos}^3 α = 3\text{ cos 30° - 4 cos}^3 30° \\[1em] = 3 \times \frac{\sqrt{3}}{2} - 4 \times \Big(\frac{\sqrt{3}}{2}\Big)^3 \\[1em] = \frac{3\sqrt{3}}{2} - 4 \times \frac{3\sqrt{3}}{8} \\[1em] = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \\[1em] = 0.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} = \frac{1 - 1}{1 + 1} \\[1em] = 0 \\[1em] = \text{tan 0°}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

⇒ sin α = $\frac{1}{2}$

⇒ sin α = sin 30°

⇒ α = 30°.

Also,

⇒ cos β = $\frac{1}{2}$

⇒ cos β = cos 60°

⇒ β = 60°.

(α + β) = 30° + 60° = 90°.

Hence, Option 4 is the correct option.

In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 90°.

⇒ cos (A + B) = cos 90° = 0.

Hence, Option 1 is the correct option.

By formula,

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

$$\begin{aligned}\Rightarrow \text{tan 30°} = \frac{AB}{BC} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{BC} \\[1em] \Rightarrow BC = 10\sqrt{3} \text{ cm}.\end{aligned}$$

Hence, Option 2 is the correct option.

By formula,

cos P = $\frac{\text{Base}}{\text{Hypotenuse}}$

$$\begin{aligned}\Rightarrow \text{cos P} = \frac{PQ}{PR} \\[1em] \Rightarrow \text{cos P} = \frac{4}{8} \\[1em] \Rightarrow \text{cos P} = \frac{1}{2} \\[1em] \Rightarrow \text{cos P} = \text{cos 60°} \\[1em] \Rightarrow P = 60°.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

sin 18° - cos 72° = 0

Solving L.H.S.,

⇒ sin (90° - 72°) - cos 72°

⇒ cos 72° - cos 72°

⇒ 0.

Since, L.H.S. = R.H.S. = 0.

∴ Statement 1 is true.

The statement sin θ = cos (90° - θ) is true.

This is a fundamental trigonometric identity, often referred to as the cofunction identity.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.

Given, tan 30° + sec 30° = cot 30°.

Solving, L.H.S.

⇒ tan 30° + sec 30°

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}\\[1em] \Rightarrow \frac{1 + 2}{\sqrt{3}}\\[1em] \Rightarrow \frac{3}{\sqrt{3}}\\[1em] \Rightarrow \sqrt{3}\end{aligned}$$

Solving, R.H.S. = cot 30° = $\sqrt{3}$

Since, L.H.S. = R.H.S.

∴ Assertion (A) is true.

According to reason (R) : sec θ = $\frac{\text{cosec θ}}{\text{cot θ}}$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow \frac{\text{cosec θ}}{\text{cot θ}}\\[1em] \Rightarrow \frac{\frac{1}{\text{sin θ}}}{\frac{\text{cos θ}}{\text{sin θ}}}\\[1em] \Rightarrow \frac{\text{sin θ}}{\text{sin θ} \times \text{cos θ}}\\[1em] \Rightarrow \frac{1}{\text{cos θ}}\\[1em] \Rightarrow \text{sec θ}.\end{aligned}$$

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Given, cos (A + B) = $\frac{1}{2}$

Since 0° < A + B ≤ 90°, the angle whose cosine = $\frac{1}{2}$ is 60°.

So, A + B = 60° ....................(1)

sin (A - B) = $\frac{1}{2}$

Since A > B, A - B will be positive. The angle whose sine = $\frac{1}{2}$ is 30°.

So, A - B = 30° ....................(2)

Adding equations (1) and (2), we get :

⇒ (A + B) + (A - B) = 60° + 30°

⇒ A + B + A - B = 90°

⇒ 2A = 90°

⇒ A = $\frac{90°}{2}$

⇒ A = 45°

Substituting the value of A in equation (1), we get :

⇒ 45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°.

∴ Assertion (A) is true.

sin 60° = $\frac{\sqrt{3}}{2}$

cos 60° = $\frac{1}{2}$

As sin 60° ≠ cos 60°

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

According to assertion, if ΔABC is equilateral, then cos A + cos B + cos C = sin A + sin B + sin C.

In an equilateral triangle, all three angles are equal to 60°. So, A = B = C = 60°.

Taking L.H.S.,

⇒ cos A + cos B + cos C

⇒ cos 60° + cos 60° + cos 60°

⇒ $\frac{1}{2} + \frac{1}{2} + \frac{1}{2}$

⇒ $\frac{3}{2}$

Taking R.H.S.,

⇒ sin A + sin B + sin C

⇒ sin 60° + sin 60° + sin 60°

⇒ $\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$

⇒ $\frac{3\sqrt{3}}{2}$

As L.H.S. ≠ R.H.S.

∴ Assertion (A) is false.

According to reason, in isosceles right angled triangle ABC, cos A + cos C = sin A + sin C.

An isosceles right-angled triangle has angles 45°, 45°, and 90°. In triangle ABC, the right angle is at B, so ∠B = 90°. Then the two equal angles are A = 45° and C = 45°.

Taking L.H.S.,

⇒ cos A + cos C

⇒ cos 45° + cos 45°

⇒ $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$

⇒ $\frac{2}{\sqrt{2}} = \sqrt{2}$

Taking R.H.S.,

⇒ sin A + sin C

⇒ sin 45° + sin 45°

⇒ $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$

⇒ $\frac{2}{\sqrt{2}} = \sqrt{2}$

As L.H.S. = R.H.S.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{\sqrt{3}}{2}\Big)^2 - \Big(\frac{1}{\sqrt{2}}\Big)^2 + 3 \Big(\frac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \frac{3}{4} - \frac{1}{2} + 3 \times \frac{1}{3} \\[1em] \Rightarrow \frac{3 - 2}{4} + 1 \\[1em] \Rightarrow \frac{1}{4} + 1 \\[1em] \Rightarrow \frac{1 + 4}{4} \\[1em] \Rightarrow \frac{5}{4} \\[1em] \Rightarrow 1\frac{1}{4}.\end{aligned}$$

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

Solving,

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

Hence, $\frac{\text{2 cos}^2 45° + \text{3 tan}^2 30°}{\sqrt{3}\text{cos 30° + \text{sin 30°}}}$ = 1.

Solving,

$$\begin{aligned}\Rightarrow \text{sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°} \\[1em] \Rightarrow \frac{2}{\sqrt{3}} \times \sqrt{3} + \text{sin 45°} \times \frac{1}{\text{sin 45°}} + \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} \\[1em] \Rightarrow 2 + 1 + \frac{1}{2} \\[1em] \Rightarrow \frac{4 + 2 + 1}{2} \\[1em] \Rightarrow \frac{7}{2} \\[1em] \Rightarrow 3\frac{1}{2}.\end{aligned}$$

Hence, sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° = $3\frac{1}{2}.$

(i) To verify,

cos⁴ A - sin⁴ A = cos 2A

Substituting value of A in L.H.S. of the above equation, we get :

$$\begin{aligned}\Rightarrow \text{cos}^4 \space A - \text{sin}^4 \space A \\[1em] \Rightarrow \text{cos}^4 \space 30° - \text{sin}^4 \space 30° \\[1em] \Rightarrow \Big(\frac{\sqrt{3}}{2}\Big)^4 - \Big(\frac{1}{2}\Big)^4 \\[1em] \Rightarrow \frac{9}{16} - \frac{1}{16} \\[1em] \Rightarrow \frac{8}{16} \\[1em] \Rightarrow \frac{1}{2}.\end{aligned}$$

Substituting value of A in R.H.S. of the above equation, we get :

⇒ cos 2A = cos 2(30°) = cos 60° = $\frac{1}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that cos⁴ A - sin⁴ A = cos 2A.

(ii) To verify,

4 cos A cos (60° - A) cos (60° + A) = cos 3A.

Substituting value of A in L.H.S. of the above equation, we get :

$$\begin{aligned}\Rightarrow \text{4 cos 30° cos (60° - 30°) cos (60° + 30°)} \\[1em] \Rightarrow \text{4 cos 30° cos 30° cos 90°} \\[1em] \Rightarrow 4 \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \times 0 \\[1em] \Rightarrow 0.\end{aligned}$$

Substituting value of A in R.H.S. of the above equation, we get :

$\Rightarrow \text{cos 3(30°)} = \text{cos 90°} = 0.$

Since, L.H.S. = R.H.S.

Hence, proved that 4 cos A cos (60° - A) cos (60° + A) = cos 3A

To verify,

$\frac{\text{sin A}}{\text{cos A + sin A sin B}} = \frac{2}{3}$.

Substituting value of A and B in L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{sin 45°}}{\text{cos 45° + sin 45° sin 30°}} \\[1em] \Rightarrow \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \times \frac{1}{2}} \\[1em] \Rightarrow \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}\Big(1 + \frac{1}{2}\Big)} \\[1em] \Rightarrow \frac{\bcancel{\frac{1}{\sqrt{2}}}}{\bcancel{\frac{1}{\sqrt{2}}}\Big(\frac{2 + 1}{2}\Big)} \\[1em] \Rightarrow \frac{1}{\frac{3}{2}} \\[1em] \Rightarrow \frac{2}{3}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A}}{\text{cos A + sin A sin B}} = \frac{2}{3}$.

(i) To verify,

$\frac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$.

Substituting value of A and B in L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{sin(A + B)}}{\text{cos A cos B}} \\[1em] \Rightarrow \frac{\text{sin(60° + 30°)}}{\text{cos 60° cos 30°}} \\[1em] \Rightarrow \frac{\text{sin 90°}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}} \\[1em] \Rightarrow \frac{1}{\frac{\sqrt{3}}{4}} \\[1em] \Rightarrow \frac{4}{\sqrt{3}}.\end{aligned}$$

Substituting value of A and B in R.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \text{tan A + tan B} \\[1em] \Rightarrow \text{tan 60° + tan 30°} \\[1em] \Rightarrow \sqrt{3} + \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \frac{3 + 1}{\sqrt{3}} \\[1em] \Rightarrow \frac{4}{\sqrt{3}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$.

(ii) To verify,

$\frac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$

Substituting value of A and B in L.H.S. of the equation, we get :

$$\begin{aligned}\Rightarrow \frac{\text{sin(A - B)}}{\text{sin A sin B}} \\[1em] \Rightarrow \frac{\text{sin(60° - 30°)}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \frac{\text{sin 30°}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2} \times \frac{1}{2}} \\[1em] \Rightarrow \frac{\frac{1}{2}}{\frac{\sqrt{3}}{4}} \\[1em] \Rightarrow \frac{4}{2\sqrt{3}} \\[1em] \Rightarrow \frac{2}{\sqrt{3}}.\end{aligned}$$

Substituting value of A and B in R.H.S. of the equation, we get :

⇒ cot B - cot A

⇒ cot 30° - cot 60°

⇒ $\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}}$

⇒ $\frac{2}{\sqrt{3}}$.

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$.

Given,

$\phantom{\Rightarrow}\sqrt{2}$ tan 2θ = $\sqrt{6}$

⇒ tan 2θ = $\frac{\sqrt{6}}{\sqrt{2}}$

⇒ tan 2θ = $\sqrt{3}$

⇒ tan 2θ = tan 60°

⇒ 2θ = 60°

⇒ θ = $\frac{60°}{2}$

⇒ θ = 30°.

Substituting value of θ in sin θ + $\sqrt{3}$ cos θ - 2 tan² θ, we get :

⇒ sin 30° + $\sqrt{3}$ cos 30° - 2 tan² 30°

$$\begin{aligned}\Rightarrow \frac{1}{2} + \sqrt{3} \times \frac{\sqrt{3}}{2} - 2 \times \Big(\frac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \frac{1}{2} + \frac{3}{2} - \frac{2}{3} \\[1em] \Rightarrow \frac{3 + 9 - 4}{6} \\[1em] \Rightarrow \frac{8}{6} \\[1em] \Rightarrow \frac{4}{3}.\end{aligned}$$

Hence, sin θ + $\sqrt{3}$ cos θ - 2 tan² θ = $\frac{4}{3}.$

Given,

⇒ (cosec 3θ – 2)(cot 2θ – 1) = 0

⇒ cosec 3θ – 2 = 0 or cot 2θ – 1 = 0

⇒ cosec 3θ = 2 or cot 2θ = 1

⇒ cosec 3θ = cosec 30° or cot 2θ = cot 45°

⇒ 3θ = 30° or 2θ = 45°

⇒ θ = $\frac{30°}{3}$ or θ = $\frac{45°}{2}$

⇒ θ = 10° or θ = $22\frac{1}{2}°$

Hence, θ = 10° or θ = $22\frac{1}{2}°$.

Given,

⇒ tan (A + B) = $\sqrt{3}$

⇒ tan (A + B) = tan 60°

⇒ A + B = 60° ..........(1)

⇒ tan (A - B) = 1

⇒ tan (A - B) = tan 45°

⇒ A - B = 45° ..........(2)

Adding (1) and (2) we get :

⇒ A + B + A - B = 60° + 45°

⇒ 2A = 105°

⇒ A = $\frac{105}{2}$

⇒ A = $52\frac{1}{2}^{\circ}$

Substituting value of A in (2) we get :

⇒ $\frac{105}{2}$ - B = 45°

⇒ B = $\frac{105}{2}$ - 45°

⇒ B = $\frac{105 - 90}{2}$

⇒ B = $\frac{15}{2}$

⇒ B = $7\frac{1}{2}^{\circ}$

Hence, A = $52\frac{1}{2}^{\circ}$ and B = $7\frac{1}{2}^{\circ}$

Solving,

⇒ sin² 28° + sin² 62° - tan² 45°

⇒ sin² 28° + sin² (90° - 28°) - tan² 45°

⇒ sin² 28° + cos² 28° - 1 [∵ sin (90 - θ) = cos θ]

⇒ 1 - 1 [∵ sin² θ + cos² θ]

⇒ 0.

Hence, sin² 28° + sin² 62° - tan² 45° = 0.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{2 cos 27°}}{\text{sin 63°}} + \frac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°} \\[1em] \Rightarrow \frac{\text{2 cos 27°}}{\text{sin (90° - 27°)}} + \frac{\text{tan 27°}}{\text{cot (90° - 27°)}} + 1 \\[1em] \Rightarrow \frac{\text{2 cos 27°}}{\text{cos 27°}} + \frac{\text{tan 27°}}{\text{tan 27°}} + 1 \\[1em] \Rightarrow 2 + 1 + 1 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\frac{\text{2 cos 27°}}{\text{sin 63°}} + \frac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°}$ = 4.

Solving,

⇒ cos 18° sin 72° + sin 18° cos 72°

[As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ]

⇒ cos 18° sin (90° - 18°) + sin 18° cos (90° - 18°)

⇒ cos 18° cos 18° + sin 18° sin 18°

⇒ sin² 18° + cos² 18°

⇒ 1.

Hence, cos 18° sin 72° + sin 18° cos 72° = 1.

Solving,

⇒ 5 sin 50° sec 40°- 3 cos 59° cosec 31°

⇒ 5 sin 50° sec (90° - 50°) - 3 cos 59° cosec (90° - 59°)

[As, sec (90 - θ) = cosec θ and cosec (90 - θ) = sec θ]

⇒ 5 sin 50° cosec 50° - 3 cos 59° sec 59°

⇒ $\text{5 sin 50°} \times \frac{1}{\text{sin 50°}} - \text{3 cos 59°} \times \frac{1}{\text{cos 59°}}$

⇒ 5 - 3

⇒ 2.

Hence, 5 sin 50° sec 40°- 3 cos 59° cosec 31° = 2.

Solving L.H.S. of above equation,

$$\begin{aligned}\Rightarrow \frac{\text{cos (90° - θ) sec (90° - θ) tan θ}}{\text{cosec (90 - θ) sin (90 - θ) cot (90 - θ)}} + \frac{\text{tan (90 - θ)}}{\text{cot θ}} \\[1em] \Rightarrow \frac{\text{sin θ. cosec θ. tan θ}}{\text{sec θ. cos θ. tan θ}} + \frac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow \frac{\text{sin θ } \times \frac{1}{\text{sin θ}}\times \text{ tan θ}}{\frac{1}{\text{cos θ}} \times \text{. cos θ. tan θ}} + \frac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos (90° - θ) sec (90° - θ) tan θ}}{\text{cosec (90 - θ) sin (90 - θ) cot (90 - θ)}} + \frac{\text{tan (90 - θ)}}{\text{cot θ}} = 2.$

(i) Given,

⇒ sin 3A = cos 2A

⇒ sin 3A = sin (90° - 2A)

⇒ 3A = 90° - 2A

⇒ 5A = 90°

⇒ A = $\frac{90}{5}$

⇒ A = 18°.

Hence, A = 18°.

(ii) Given,

⇒ tan 5A = cot A

⇒ tan 5A = tan (90° - A)

⇒ 5A = 90° - A

⇒ 6A = 90°

⇒ A = $\frac{90}{6}$

⇒ A = 15°.

Hence, A = 15°.

(i) Given,

⇒ sin (θ + 36°) = cos θ

⇒ sin (θ + 36°) = sin (90° - θ)

⇒ θ + 36° = 90° - θ

⇒ 2θ = 90° - 36°

⇒ 2θ = 54°

⇒ θ = $\frac{54}{2}$

⇒ θ = 27°.

Hence, θ = 27°.

(ii) Given,

⇒ sec 4θ = cosec (θ - 20°)

⇒ sec 4θ = sec [90° - (θ - 20°)]

⇒ 4θ = [90° - (θ - 20°)]

⇒ 4θ = 110° - θ

⇒ 5θ = 110°

⇒ θ = $\frac{110}{5}$

⇒ θ = 22°.

Hence, θ = 22°.

In △ABC,

tan 30° = $\frac{\text{Perpendicular}}{\text{Base}}$

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{AB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}}{AB} \\[1em] \Rightarrow AB = 6 \text{ cm}.\end{aligned}$$

In △ABE,

90° = 30° + ∠ABE [As, exterior angle is equal to sum of two opposite interior angles]

∠ABE = 90° - 30° = 60°.

From figure,

∠ABD = ∠ABE = 60°

In △ABD,

tan ∠ABD = tan 60° = $\frac{\text{Perpendicular}}{\text{Base}}$

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{AD}{AB} \\[1em] \Rightarrow AD = AB\sqrt{3} \\[1em] \Rightarrow AD = 6\sqrt{3} \text{ cm}.\end{aligned}$$

Hence, AD = $6\sqrt{3}$ cm.