Prove the following:
$ 1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha $

To do:

We have to prove that \( 1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=1+\frac{\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}}{1+\frac{1}{\sin \alpha}}$

$=1+\frac{\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}}{\frac{\sin \alpha+1}{\sin \alpha}}$

$=1+\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha} \times \frac{\sin \alpha}{\sin \alpha+1}$

$=1+\frac{\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$

$=\frac{\sin \alpha(\sin \alpha+1)+\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$

$=\frac{\sin^2 \alpha+\sin \alpha+\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$

$=\frac{1+\sin \alpha}{\sin \alpha(\sin \alpha+1)}$

$=\frac{1}{\sin \alpha}$

$=\operatorname{cosec} \alpha$

Hence proved.