Q. $cosec\ 15{^\circ} + \sec{15{^\circ}}$ is equal to

A) $2\sqrt{2}$
B) $\sqrt{6}$
C) $2\sqrt{6}$
D) $\sqrt{6} + \sqrt{2}$
Explanation

(c)
$cosec\ 15{^\circ} + \sec{15{^\circ} = \frac{2\left( \sin{15{^\circ} + \cos{15{^\circ}}} \right)}{2\sin{15{^\circ}\cos{15{^\circ}}}}}$
$= 2\left\lbrack \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \right\rbrack/\ \ \sin\ 30{^\circ}$
$= \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.