Question

If $0<x<\frac{\pi }{2}$, and if $\frac{y+1}{1+y}=\sqrt{\frac{1+\sin x}{1-\sin x}},$ then $y$ is equal to

• A. $\cot \frac{x}{2}$
• B. $\tan \frac{x}{2}$
• C. $\cot \frac{x}{2}+\tan \frac{x}{2}$
• D. $\cot \frac{x}{2}-\tan \frac{x}{2}$

Answer 1

Answer: (A)

$\cot \frac{x}{2}$

Given $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$

$\frac{1+y}{1-y}=\sqrt{\frac{(1+\sin x{)}^2}{1-{\sin }^{2}x}}$

$\frac{1+y}{1-y}=\sqrt{(\frac{1+\sin x}{\cos x}{)}^2}$

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

Applying componendo and dividendo rule

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

$\frac{1}{y}=\frac{2{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{\cos \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})}{\sin \frac{x}{2}(\sin \frac{x}{2}+\cos \frac{x}{2})}$

$⟹y=\cot \frac{x}{2}$