Answer
As left side transforms into right side, hence given identity-
$ \frac{\sin x + 1}{\cos x + \cot x}$ = $\tan x $ is true.
Work Step by Step
Given identity is-
$ \frac{\sin x + 1}{\cos x + \cot x}$ = $\tan x $
Taking L.S.
$ \frac{\sin x + 1}{\cos x + \cot x}$
= $ \frac{\sin x + 1}{\cos x + \frac{\cos x}{\sin x}}$
{Using ratio identity for $\cot x$}
= $ \frac{\sin x + 1}{\cos x. \frac{\sin x}{\sin x} + \frac{\cos x}{\sin x}}$
= $ \frac{\sin x + 1}{ \frac{\cos x \sin x}{\sin x} + \frac{\cos x}{\sin x}}$
= $ \frac{\sin x + 1}{ \frac{\cos x \sin x + \cos x}{\sin x}}$
= $ \frac{\sin x + 1}{ \frac{\cos x (\sin x + 1)}{\sin x}}$
= $ \frac{(\sin x + 1) \sin x}{\cos x (\sin x + 1)}$
=$ \frac{\sin x}{\cos x}$
= $\tan x $
= R.S.
As left side transforms into right side, hence given identity-
$ \frac{\sin x + 1}{\cos x + \cot x}$ = $\tan x $ is true.
