Question
Prove the following trigonometric identities.
$\frac{\sin\theta}{1-\cos\theta}=\text{cosec}\theta+\cot\theta$
$\frac{\sin\theta}{1-\cos\theta}=\text{cosec}\theta+\cot\theta$
✓
Answer
$\text{L.H.S}=\frac{\sin\theta}{1-\cos\theta}$
Rationalizer both $Nr$ and $Or$ with $1+\cos\theta$
$\Rightarrow\ \frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}$
$\Rightarrow\ \frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}\ [\because(\text{a}-\text{b})(\text{a}+\text{b})=\text{a}^2-\text{b}^2]$
$\Rightarrow \frac{\sin\theta+\sin\theta\cos\theta}{\sin^2\theta}\ [\because 1-\cos^2\theta-\sin^2\theta]$
$\Rightarrow \frac{\sin\theta}{\sin^2\theta}+\frac{\sin\theta\cos\theta}{\sin^2\theta}$
$\Rightarrow\ \frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$\Rightarrow\ \text{cosec }\theta+\cot\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Rationalizer both $Nr$ and $Or$ with $1+\cos\theta$
$\Rightarrow\ \frac{\sin\theta}{1-\cos\theta}\times\frac{1+\cos\theta}{1+\cos\theta}$
$\Rightarrow\ \frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}\ [\because(\text{a}-\text{b})(\text{a}+\text{b})=\text{a}^2-\text{b}^2]$
$\Rightarrow \frac{\sin\theta+\sin\theta\cos\theta}{\sin^2\theta}\ [\because 1-\cos^2\theta-\sin^2\theta]$
$\Rightarrow \frac{\sin\theta}{\sin^2\theta}+\frac{\sin\theta\cos\theta}{\sin^2\theta}$
$\Rightarrow\ \frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$\Rightarrow\ \text{cosec }\theta+\cot\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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