Question:
$\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$
Solution:
$\frac{1+\cos \theta}{1-\cos \theta}$
$=\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}$
$=\frac{(1+\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}$
$=\frac{(1+\cos \theta)^{2}}{1-\cos ^{2} \theta} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta} \quad\left(\sin ^{2} \theta+\cos ^{2} \theta=1\right)$
$=\left(\frac{1+\cos \theta}{\sin \theta}\right)^{2}$
$=\left(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\right)^{2}$
$=(\operatorname{cosec} \theta+\cot \theta)^{2}$