Question

$\sqrt{{\sec }^2\theta +{\csc }^2\theta }=\tan \theta +\cot \theta$

• A. True
• B. False

Answer 1

The correct option is A True

$\sqrt{{\text{sec}}^2\theta +{\text{cosec}}^2\theta }=\sqrt{1+{\tan }^2\theta +1+{\cot }^2\theta }$

$=\sqrt{(\tan \theta {)}^2+(\cot \theta {)}^2+2}$

$=\sqrt{(\tan \theta {)}^2+(\cot \theta {)}^2+2\left(\tan \theta \right)\left(\cot \theta \right)}$

$=\sqrt{(\tan \theta +\cot \theta {)}^2}$

$=\tan \theta +\cot \theta$

So the relation is $\text{True}$