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