Question
Prove the following trigonometric identities.
$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\Big(\frac{1-\tan\theta}{1-\cot\theta}\Big)^2=\tan^2\theta$

Answer

$\text{L.H.S}\Rightarrow\ \frac{1+\tan^2\theta}{1+\cot^2\theta}=\frac{\sec^2\theta}{\text{cosec}^2\theta}$
$\big[\because \tan^2\theta+1=\sec^2\theta,\ 1+\cot^2\theta=\text{cosec}^2\theta\big]$
$=\frac{1}{\cos^2\theta\times1}\ \sin^2=\tan^2\theta$
$\Rightarrow\ \Big[\frac{1-\tan\theta}{1-\cot\theta}\Big]^2\Rightarrow\ \Bigg[\frac{1-\tan\theta}{1-\frac{1}{\tan\theta}}\Bigg]^2$
$\Rightarrow\ \Big[\frac{1-\tan\theta}{(1-\tan\theta)}\times\tan\theta\Big]^2=\tan^2\theta$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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