Question
Prove the following identities:
$\frac{\big(1+\tan^2\theta\big)\cot\theta}{\text{cosec}^2\theta}=\tan\theta$
$\frac{\big(1+\tan^2\theta\big)\cot\theta}{\text{cosec}^2\theta}=\tan\theta$
✓
Answer
$\text{L.H.S.}=\frac{\big(1+\tan^2\theta\big)\cot\theta}{\text{cosec}^2\theta}=\frac{\sec^2\theta\cot\theta}{\text{cosec}^2\theta}$
$=\frac{1}{\cos^2\theta}\times\frac{\cos\theta}{\sin\theta}\times\sin^2\theta=\frac{\sin\theta}{\cos\theta}$ $\Big[\because\big(1+\tan^2\theta\big)=\sec^2\theta\Big]$
$=\tan\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
$=\frac{1}{\cos^2\theta}\times\frac{\cos\theta}{\sin\theta}\times\sin^2\theta=\frac{\sin\theta}{\cos\theta}$ $\Big[\because\big(1+\tan^2\theta\big)=\sec^2\theta\Big]$
$=\tan\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
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