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

Trigonometry (10)

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

Answer

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

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

$$ = \tan\theta + \cot\theta = (RHS) $$