Question

$\frac{\cos x}{\cos x-2y}=\lambda \Rightarrow \tan (x-y)\tan y$ is equal to

• A. $\frac{1+\lambda }{1-\lambda }$
• B. $\frac{1-\lambda }{1+\lambda }$
• C. $\frac{\lambda }{1+\lambda }$
• D. $\frac{\lambda }{1-\lambda }$

Answer 1

The correct option is B $\frac{1-\lambda }{1+\lambda }$

Given: $\frac{cosx}{cos(x-2y)}=\lambda$

Consider, $tan(x-y)tany$

$=\frac{sin(x-y)}{cos(x-y)}\frac{siny}{cosy}$

$=\frac{2sin(x-y)}{2cos(x-y)}\frac{siny}{cosy}$

$=\frac{cos(x-y-y)-cos(x-y+y)}{cos(x-y+y)+cos(x-y-y)}$[$\because 2sinAsinB=cos(A-B)-cos(A+B),2cosAcosB=cos(A+B)+cos(A-B)$]

$=\frac{cos(x-2y)-cosx}{cosx+cos(x-2y)}$

$=\frac{1-\frac{cosx}{cos(x-2y)}}{\frac{cosx}{cos(x-2y)}+1}$

$=\frac{1-\lambda }{\lambda +1}$

$=\frac{1-\lambda }{1+\lambda }$

Hence, $\text{B}$ is the correct option.