Question

Prove that: $\frac{1}{\sin (x-a)\sin (x-b)}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$

Answer 1

To Prove : $\frac{1}{\sin (x-a)\sin (x-b)}=\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$

Taking R.H.S.

$\frac{\cot (x-a)-\cot (x-b)}{\sin (a-b)}$

$\frac{\frac{\cos (x-a)}{\sin (x-a)}\frac{\cos (x-b)}{\sin (x-b)}}{\sin (a-b)}$

$=\frac{\sin (x-b)\cos (x-a)-\sin (x-a)\cos (x-b)}{\sin (x-a)\sin (x-b)\sin (a-b)}$

$=\frac{\sin (x-b-x+a)}{\sin (x-a)\sin (x-b)\sin (a-b)}$
$[\because \sin A\cos B-\sin B\cos A]$

$=\frac{1}{\sin (x-a)\sin (x-b)}$

Hence proved.