Question

Prove that $\sqrt{{\sec }^2\theta +{\csc }^2\theta }=\tan \theta +\cot \theta$if $\theta$ lies in the first quadrant.

Answer 1

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

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

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

hence proved