Prove that,
$1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha} = \operatorname{cosec} \alpha$

(N/A) $L$.$H$.$S$. $= 1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha}$
$= 1 + \frac{\cos^{2} \alpha / \sin^{2} \alpha}{1 + 1 / \sin \alpha} \quad [\because \cot \theta = \frac{\cos \theta}{\sin \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= 1 + \frac{\cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)} = \frac{\sin \alpha(1 + \sin \alpha) + \cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)}$
$= \frac{\sin \alpha + (\sin^{2} \alpha + \cos^{2} \alpha)}{\sin \alpha(1 + \sin \alpha)} \quad [\because \sin^{2} \theta + \cos^{2} \theta = 1]$
$= \frac{\sin \alpha + 1}{\sin \alpha(1 + \sin \alpha)} = \frac{1}{\sin \alpha} \quad [\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= \operatorname{cosec} \alpha = \text{R.H.S.}$