Question
Prove the following.
$\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta}=2\text{cosec}\theta$
$\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta}=2\text{cosec}\theta$
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Answer
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)}\ [\because(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}]$
$=\frac{1+1+2\cos\theta}{\sin\theta(1+\cos\theta)}\ [\because\sin^2\theta+\cos^2\theta=1]$
$=\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{2}{\sin\theta}$
$=2\text{cosec}\theta=$ RHS $\Big[\because\ \text{cosec}\theta=\frac{1}{\sin\theta}\Big]$
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