Question

Prove that: $\left(ii\right)$ $\frac{1}{\sin (x-a)\cos (x-b)}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$

Answer 1

To Prove : $\frac{1}{\sin (x-a)\cos (x-b)}=\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$
Taking R.H.S.
$\frac{\cot (x-a)+\tan (x-b)}{\cos (a-b)}$
$\frac{\frac{\cos (x-a)}{\sin (x-a)}+\frac{\sin (x-b)}{\cos (x-b)}}{\cos (a-b)}$
$=\frac{\cos (x-b)\cos (x-a)+\sin (x-b)\sin (x-a)}{\sin (x-a)\cos (x-b)\cos (a-b)}$
$[\because \cos A\cos B+\sin A\sin B=\cos (A-B\left)\right]$
$=\frac{\cos (x-b-x+a)}{\sin (x-a)\cos (x-b)\cos (a-b)}$
$=\frac{1}{\sin (x-a)\cos (x-b)}$
Hence proved.