Prove the following trigonometry identity :
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ.sec θ – 2 tan θ
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Solving,
(sin θ + cos θ)(cosec θ - sec θ)
$$\begin{aligned}\Rightarrow (\sin θ + \cos θ) \times \Big(\frac{1}{\sin θ} - \frac{1}{\cos θ}\Big) \\[1em] \Rightarrow (\sin θ + \cos θ) \times \Big(\frac{\cos θ - \sin θ }{\sin θ \cos θ}\Big) \\[1em] \Rightarrow \frac{\cos^2 θ - \sin^2 θ }{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1- \sin^2 θ - \sin^2 θ}{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1 - 2\sin^2 θ }{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1}{\sin θ. \cos θ} - \frac{2 \sin^2 θ}{\sin θ.\cos θ} \\[1em] \Rightarrow \cosec θ.\sec θ - \frac{2 \sin θ}{\cos θ} \\[1em] \Rightarrow \cosec θ.\sec θ - 2\tan θ \\[1em]\end{aligned}$$
Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.