Question

If $0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}$, and if $\frac{y+1}{1-y}=\sqrt{\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}}$, then y is equal to
(a) $\cot \frac{\mathrm{\theta }}{2}$
(b) $\tan \frac{\mathrm{\theta }}{2}$
(c) $\cot \frac{\mathrm{\theta }}{2}+\tan \frac{\mathrm{\theta }}{2}$
(d) $\cot \frac{\mathrm{\theta }}{2}-\tan \frac{\mathrm{\theta }}{2}$

Answer 1

(b) $\tan \frac{\mathrm{\theta }}{2}$

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