Question
Prove the following trigonometric identities.
$1+\frac{\cot^2\theta}{1+\text{cosec }\theta}=\text{cosec }\theta$

Answer

$\text{L.H.S}=1+\frac{\cot^2\theta}{1+\text{cosec }\theta}$
$1+\frac{\text{cosec}^2\theta-1}{1+\text{cosec }\theta}\ \big[\because \text{cosec}^2\theta-\cot^2\theta=1,\cot^2\theta=\text{cosec }^2\theta-1\big]$
$1+\frac{(\text{cosec }\theta-1)(\text{cosec }\theta+1)}{1+\text{cosec }\theta}$
$=1+\text{cosec }\theta-1\ \big[\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2\text{a}=\text{cosec}^2\theta-1\big]$
$=\text{cosec }\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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