Question

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

Answer 1

$\sqrt{se{c}^2\theta +cose{c}^2\theta }=\sqrt{1+ta{n}^2\theta +1+co{t}^2\theta }$

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

$=\sqrt{ta{n}^2\theta +co{t}^2\theta +2.tan\theta .cot\theta }[\because cot\theta \frac{1}{tan\theta }]$

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

$=\sqrt{tan\theta +cot\theta }$