Prove the following identity:

$\Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) = 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big)$

L.H.S. of the equation can be written as,

$\Rightarrow \Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) \\[1em] \Rightarrow \dfrac{\sin A}{\dfrac{\cos A}{\sin A} + \dfrac{1}{\sin A}} \\[1em] \Rightarrow \dfrac{\sin A}{\dfrac{\cos A + 1}{\sin A}} \\[1em] \Rightarrow \dfrac{\sin^2 A}{\cos A + 1} \\[1em]$

Multiplying numerator and denominator by (cos A - 1), we get :

$\Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos A + 1(\cos A − 1)} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos^2 A - 1} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{-\sin^2 A} \\[1em] \Rightarrow -(\cos A − 1) \\[1em] \Rightarrow 1 - \cos A$

R.H.S. of the equation can be written as,

$\Rightarrow 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big) \\[1em] \Rightarrow 2 + \dfrac{\sin A}{\dfrac{\cos A}{\sin A} - \dfrac{1}{\sin A}} \\[1em] \Rightarrow 2 + \dfrac{\sin A}{\dfrac{\cos A - 1}{\sin A}} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A}{\cos A - 1} \\[1em]$

Multiplying numerator and denominator by (cos A + 1), we get :

$\Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos A - 1(\cos A + 1)} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos^2 A - 1} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{-\sin^2 A} \\[1em] \Rightarrow 2 -(\cos A + 1) \\[1em] \Rightarrow 1 - \cos A$

Since, L.H.S. = R.H.S.

Hence, proved that $\Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) = 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big)$.