Question

Prove :

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

Answer 1

LHS :

$\sqrt{se{c}^2\theta +cose{c}^2\theta }$

$=\sqrt{(1+ta{n}^2\theta )+(1+co{t}^2\theta )}$

$=\sqrt{2+ta{n}^2\theta +co{t}^2\theta }$

$=\sqrt{ta{n}^{\theta }+co{t}^2\theta +2\tan \theta \cot \theta }$

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

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