Q. $\frac{\cos x}{\cos{(x - 2y)}} = \lambda\ \Rightarrow \tan{(x - y)}\tan y$ is equal to
A) $\frac{1 + \lambda}{1 - \lambda}$
B) $\frac{1 - \lambda}{1 + \lambda}$
C) $\frac{\lambda}{1 + \lambda}$
D) $\frac{\lambda}{1 - \lambda}$
Explanation
(b)
Now, $\tan{(x - y})\tan{y = \frac{\sin(x - y)\sin y}{\cos{(x - y)\cos y}} \times \frac{2}{2}}$
$= \frac{\cos{(x - 2y) - \cos{(x)}}}{\cos{(x - 2y) + \cos(x)}} = \frac{1 - \frac{\cos x}{\cos{(x - 2y)}}}{1 + \frac{\cos{(x)}}{\cos{(x - 2y)}}}$
$= \frac{1 - \lambda}{1 + \lambda}$ $\left\lbrack Given,\ \lambda = \frac{\cos x}{\cos{(x - 2y)}}\ \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.