Question
Prove the following trigonometric identities.
$\tan\theta+\frac{1}{\tan\theta}=\sec\theta\text{cosec}\theta$
$\tan\theta+\frac{1}{\tan\theta}=\sec\theta\text{cosec}\theta$
✓
Answer
We know that, $\sec^2\theta-\tan^2\theta=1$
So,
$\text{L.H.S.} = \tan\theta+\frac{1}{\tan\theta}$
$\tan\theta+\frac{1}{\tan\theta}=\frac{\tan^2\theta+1}{\tan\theta}$
$=\frac{\sec^2\theta}{\tan\theta}$
$=\sec\theta\frac{\sec\theta}{\tan\theta}$
$=\sec\theta\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=\sec\theta\frac{1}{\sin\theta}$
$=\sec\theta\text{cosec}\theta$
$= \text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
So,
$\text{L.H.S.} = \tan\theta+\frac{1}{\tan\theta}$
$\tan\theta+\frac{1}{\tan\theta}=\frac{\tan^2\theta+1}{\tan\theta}$
$=\frac{\sec^2\theta}{\tan\theta}$
$=\sec\theta\frac{\sec\theta}{\tan\theta}$
$=\sec\theta\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=\sec\theta\frac{1}{\sin\theta}$
$=\sec\theta\text{cosec}\theta$
$= \text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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