Question

\[ \text{If } 0

\[ \frac{y+1}{1-y} = \sqrt{\frac{1+\sin x}{1-\sin x}} \]

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) \(\cot \frac{x}{2}-\tan \frac{x}{2}\)

Solution

\[ \sqrt{\frac{1+\sin x}{1-\sin x}} = \frac{1+\sin x}{\sqrt{1-\sin^2 x}} \]

\[ = \frac{1+\sin x}{\cos x} \]

\[ = \sec x+\tan x \]

Using identity,

\[ \sec x+\tan x = \tan\left(\frac{\pi}{4}+\frac{x}{2}\right) \]

\[ = \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \]

Comparing with

\[ \frac{y+1}{1-y} \]

we get

\[ y=\tan \frac{x}{2} \]

Answer

\[ \boxed{\tan \frac{x}{2}} \]

Correct Option: (b)