Question:
Prove the following identities (1-16)
$\operatorname{cosec} x(\sec x-1)-\cot x(1-\cos x)=\tan x-\sin x$
Solution:
$\mathrm{LHS}=\operatorname{cosec} x(\sec x-1)-\cot x(1-\cos x)$
$=\frac{1}{\sin x}\left(\frac{1}{\cos x}-1\right)-\frac{\cos x}{\sin x}(1-\cos x)$
$=\frac{1}{\sin x}\left(\frac{1-\cos x}{\cos x}\right)-\frac{\cos x}{\sin x}(1-\cos x)$
$=\left(\frac{1-\cos x}{\sin x}\right)\left(\frac{1}{\cos x}-\cos x\right)$
$=\left(\frac{1-\cos x}{\sin x}\right)\left(\frac{1-\cos ^{2} x}{\cos x}\right)$
$=\left(\frac{1-\cos x}{\sin x}\right)\left(\frac{\sin ^{2} x}{\cos x}\right)$
$=(1-\cos x)\left(\frac{\sin x}{\cos x}\right)$
$=\frac{\sin x}{\cos x}-\sin x$
$=\tan x-\sin x$
$=\mathrm{RHS}$
Hence proved.