Question
Prove the following trigonometric identities.
$\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta$
$\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta$
✓
Answer
We have to prove $\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta$
We know that, $\sec^2\theta-\tan^2\theta=1$
So,
$\text{L.H.S}=\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\frac{(1+\cot^2\theta)\tan\theta}{1+\tan^2\theta}$
$=\frac{\Big(1+\frac{1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{\Big(\frac{\tan^2\theta+1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{(1+\tan^2\theta)\tan\theta}{\tan^2\theta(1+\tan^2\theta)}$
$=\frac{1}{\tan\theta}$
$=\cot\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
We know that, $\sec^2\theta-\tan^2\theta=1$
So,
$\text{L.H.S}=\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\frac{(1+\cot^2\theta)\tan\theta}{1+\tan^2\theta}$
$=\frac{\Big(1+\frac{1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{\Big(\frac{\tan^2\theta+1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{(1+\tan^2\theta)\tan\theta}{\tan^2\theta(1+\tan^2\theta)}$
$=\frac{1}{\tan\theta}$
$=\cot\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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