Question

If $\cos x=\frac{2\cos y-1}{2-\cos y}$, where $x,yϵ(0,\pi )$, then $\tan \frac{x}{2}.\cot \frac{y}{2}$ is equal to-

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (A)

$\sqrt{2}$

Given $\cos x=\frac{2\cos y-1}{2-\cos y}$

Applying compoundo and dividendo rule

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

$⟹\frac{2{\cos }^{2}x/2}{2{\sin }^{2}x/2}=\frac{1+\cos y}{3(1-\cos y)}=\frac{{\cos }^{2}y/2}{3\left(2{\sin }^{2}y/2\right)}$

$⟹{\tan }^2\frac{x}{2}=3{\tan }^2\frac{y}{2}$

$⟹\tan \frac{x}{2}\cot \frac{y}{2}=\sqrt{3}$