Question

If $\cos x=k\cos (x-2y),$ then $\tan (x-y)\tan y$ is equal to

• A. $\frac{1+k}{1-k}$
• B. $\frac{1-k}{1+k}$
• C. $\frac{2k}{k+1}$
• D. $\frac{k-1}{2k+1}$

Answer 1

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

$\cos x=k\cos (x-2y)$

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

Applying componendo-Dividendo, we get

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

$\frac{2\cos \left(\frac{x+x-2y}{2}\right)\cos \left(\frac{2y}{2}\right)}{2\sin \left(\frac{x+x-2y}{2}\right)\sin \left(\frac{x-2y-x}{2}\right)}=\frac{k+1}{k-1}$

$=\frac{\cos (x-y)\cos y}{-\sin (x-y)\sin y}=\frac{k+1}{k-1}$

$\frac{\sin (x-y)\sin y}{\cos (x-y)\cos y}=\frac{1-k}{1+k}$

$\tan (x-y).\tan y=\frac{1-k}{1+k}$.