Q. If $\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \frac{x}{y}$, then the value of $\tan A$ is equal to
A) $\frac{x^{2} + y^{2}}{x^{2} - y^{2}}$
B) $\frac{2xy}{x^{2} + y^{2}}$
C) $\frac{2xy}{x^{2} - y^{2}}$
D) $\frac{2xy}{y^{2} - x^{2}}$
Explanation
(c)
$\sqrt{\frac{1 + \cos A}{1 - \cos A}} = \sqrt{\frac{2\cos^{2}\frac{A}{2}}{2\sin^{2}\frac{A}{2}} =}\frac{x}{y}$
$\Rightarrow \ \tan{\frac{A}{2} = \frac{y}{x}}$
Now, $\tan{A = \frac{2\tan\frac{A}{2}}{1 - \tan^{2}\frac{A}{2}}}$
$= \frac{2xy}{x^{2} - y^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.