Question

If θ lies in Quadrant 1, $\sqrt{se{c}^2\theta +cose{c}^2\theta }$ =

• A. tan θ + cot θ
• B. tan θ - cot θ
• C. - tan θ + cot θ
• D. - tan θ - cot θ

Answer 1

The correct option is A

tan θ + cot θ

LHS = $\sqrt{se{c}^2\theta +cose{c}^2\theta }$ = $\sqrt{(1+ta{n}^2\theta )+(1+co{t}^2\theta )}$ = $\sqrt{ta{n}^2\theta +co{t}^2\theta +2tan\theta cot\theta }$

= $\sqrt{(tan\theta +cot\theta {)}^2}$ = $tan\theta +cot\theta$ = R.H.S

Hence,proved.