Question:
Prove the following trigonometric identities.
$\tan \theta+\frac{1}{\tan \theta}=\sec \theta \operatorname{cosec} \theta$
Solution:
We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$
So,
$\tan \theta+\frac{1}{\tan \theta}=\frac{\tan ^{2} \theta+1}{\tan \theta}$
$=\frac{\sec ^{2} \theta}{\tan \theta}$
$=\sec \theta \frac{\sec \theta}{\tan \theta}$
$=\sec \theta \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$
$=\sec \theta \frac{1}{\sin \theta}$
$=\sec \theta \operatorname{cosec} \theta$
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