Question

If $0<x<\frac{\mathrm{\pi }}{2}$, 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

(b) $\tan \frac{x}{2}$

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