Answer
As left side transforms into right side, hence given identity-
$ \frac{1}{\csc x - \cot x}$ = $\csc x + \cot x $ is true.
Work Step by Step
Given identity is-
$ \frac{1}{\csc x - \cot x}$ = $\csc x + \cot x $
Taking L.S.
$ \frac{1}{\csc x - \cot x}$
= $\frac{1}{(\csc x - \cot x)} . \frac{(\csc x + \cot x)}{(\csc x + \cot x)} $
{Multiplying and dividing by , $(\csc x + \cot x)$}
= $\frac{(\csc x + \cot x)}{(\csc x - \cot x)(\csc x + \cot x)} $
= $\frac{(\csc x + \cot x)} { \csc^{2} x - \cot^{2} x}$
{Recall $(a-b)(a+b) $ = $a^{2} - b^{2}$ }
=$ \frac{(\csc x + \cot x)}{1}$
( From third Pythagorean identity, $\csc^{2} x - \cot^{2} x$ = $1$)
= $\csc x + \cot x $
= R.S.
As left side transforms into right side, hence given identity-
$ \frac{1}{\csc x - \cot x}$ = $\csc x + \cot x $ is true.
