Question

Prove that $\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$.

Answer 1

We use the following trigonometric identities:

${\sec }^2\theta ={\tan }^2\theta +1$ and

$cose{c}^2\theta ={\cot }^2\theta +1$

On adding these, we get:

${\sec }^2\theta +cose{c}^2\theta ={\tan }^2\theta +{\cot }^2\theta +2$

$\Rightarrow {\sec }^2\theta +cose{c}^2\theta ={\tan }^2\theta +{\cot }^2\theta +2\tan \theta \cot \theta =(\tan \theta +\cot \theta {)}^2$

$\Rightarrow \sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$

Hence Proved.