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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.