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

विकल्प

• \[\cot\frac{x}{2}\]

   

• \[\tan\frac{x}{2}\]

   

• \[\cot\frac{x}{2} + \tan\frac{x}{2}\]

   

• \[\cot\frac{x}{2} - \tan\frac{x}{2}\]

   

Answer 1

\[\tan\frac{x}{2}\]
We have: 
\[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]
\[ \Rightarrow \frac{y + 1}{1 - y} = \sqrt{\frac{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2}}}\]
\[ \Rightarrow \frac{y + 1}{1 - y} = \sqrt{\frac{\left( cos\frac{x}{2} + \sin\frac{x}{2} \right)^2}{\left( cos\frac{x}{2} - \sin\frac{x}{2} \right)^2}}\]
\[ \Rightarrow \frac{y + 1}{1 - y} = \frac{\left( cos\frac{x}{2} + \sin\frac{x}{2} \right)}{\left( cos\frac{x}{2} - \sin\frac{x}{2} \right)} \left[ \because 0 < x < \frac{\pi}{2} \Rightarrow 0 < \frac{x}{2} < \frac{\pi}{4}, 0\text{ to }\frac{\pi}{4} \cos x\text{ is greater than }\sin x \right]\]
\[ \Rightarrow \frac{y + 1}{1 - y} = \frac{\frac{cos\frac{x}{2}}{cos\frac{x}{2}} + \frac{\sin\frac{x}{2}}{cos\frac{x}{2}}}{\frac{cos\frac{x}{2}}{cos\frac{x}{2}} - \frac{\sin\frac{x}{2}}{cos\frac{x}{2}}} \]
\[ \Rightarrow \frac{1 + y}{1 - y} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} \]
Comparing both the sides: 
\[y = \tan\frac{x}{2}\]