Q. $\frac{\tan A}{1 + \sec A} + \frac{1 + \sec A}{\tan A}$ is equal to
A) $2\sin A$
B) $2\cos A$
C) $2\ cosec\ A$
D) $2\sec A$
Explanation
(c)
$\frac{\tan A}{1 + \sec A} + \frac{1 + \sec A}{\tan A} = \frac{\tan^{2}{A + 1 + \sec^{2}{A + 2\sec A}}}{\tan{A(1 + \sec A)}}$
$= \frac{2\sec^{2}{A + 2\sec A}}{\tan{A(1 + \sec A)}} = \frac{2\sec A}{\tan A}$
$= \frac{2\cos A}{\cos{A.\sin A}} = 2cosec\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.