Prove the following:
$ \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta $

Given:

\( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta \)

To do:

We have to prove that \( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta \)

Solution:

We know that,

$(a+b)^2=a^2+2ab+b^2$

$\sin ^{2} \theta+\cos ^{2} \theta=1$

Therefore,

LHS $=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$

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

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

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

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

$=\frac{2}{\sin \theta}$

$=2 \operatorname{cosec} \theta$

$=$ RHS

Hence proved.