Question
Prove the following identities:
$\frac{\sin\theta}{(1+\cos\theta)}+\frac{(1+\cos^2\theta)}{\sin\theta}=2\text{cosec}\theta$
$\frac{\sin\theta}{(1+\cos\theta)}+\frac{(1+\cos^2\theta)}{\sin\theta}=2\text{cosec}\theta$
✓
Answer
$\text{L.H.S.}=\frac{\sin\theta}{(1+\cos\theta)}+\frac{(1+\cos^2\theta)}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1-\cos\theta+1+\cos\theta}{\sin\theta}$
$=\frac{2}{\sin\theta}$
$=2\text{cosec}\theta$
$=\text{R.H.S.}$
$=\frac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{(1-\cos\theta)}{\sin\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1-\cos\theta+1+\cos\theta}{\sin\theta}$
$=\frac{2}{\sin\theta}$
$=2\text{cosec}\theta$
$=\text{R.H.S.}$
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