Prove the following identities
\[cosec x \left( \sec x - 1 \right) - \cot x \left( 1 - \cos x \right) = \tan x - \sin x\]
Solution
\[\text{ LHS }= cosec x \left( \sec x - 1 \right) - \cot x \left( 1 - \cos x \right)\]
\[ = \frac{1}{\sin x} \left( \frac{1}{\cos x} - 1 \right) - \frac{\cos x}{\sin x} \left( 1 - \cos x \right)\]
\[ = \frac{1}{\sin x} \left( \frac{1 - \cos x}{\cos x} \right) - \frac{\cos x}{\sin x} \left( 1 - \cos x \right)\]
\[ = \left( \frac{1 - \cos x}{\sin x} \right)\left( \frac{1}{\cos x} - \cos x \right)\]
\[ = \left( \frac{1 - \cos x}{\sin x} \right)\left( \frac{1 - \cos^2 x}{\cos x} \right)\]
\[ = \left( \frac{1 - \cos x}{\sin x} \right)\left( \frac{\sin^2 x}{\cos x} \right)\]
\[ = \left( 1 - \cos x \right)\left( \frac{\sin x}{\cos x} \right)\]
\[ = \frac{\sin x}{\cos x} - \sin x\]
\[ = \tan x - \sin x\]
= RHS
Hence proved.
