Prove the following trigonometry identity :

Solving,

$$\begin{aligned}\phantom{\Rightarrow} \text{(sin θ + cos θ)(cosec θ - sec θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\frac{1}{\text{sin θ}} - \frac{1}{\text{cos θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\frac{\text{cos θ - sin θ}}{\text{sin θ cos θ}}\Big) \\[1em] \Rightarrow \frac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \frac{\text{1 - 2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \quad [\because \text{cos}^2 \text{ θ} = 1 - \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow \frac{1}{\text{sin θ.cos θ}} - \frac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ} - \frac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ - 2 tan θ}.\end{aligned}$$

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.