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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) $\mathrm{cot}\frac{\mathrm{\theta }}{2}$ (b) $\mathrm{tan}\frac{\mathrm{\theta }}{2}$ (c) $\mathrm{cot}\frac{\mathrm{\theta }}{2}+\mathrm{tan}\frac{\mathrm{\theta }}{2}$ (d) $\mathrm{cot}\frac{\mathrm{\theta }}{2}-\mathrm{tan}\frac{\mathrm{\theta }}{2}$

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Solution

## (b) $\mathrm{tan}\frac{\mathrm{\theta }}{2}$ $\mathrm{We}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\frac{y+1}{1-y}=\sqrt{\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}}\phantom{\rule{0ex}{0ex}}⇒\frac{y+1}{1-y}=\sqrt{\frac{{\mathrm{cos}}^{2}\frac{\theta }{2}+{\mathrm{sin}}^{2}\frac{\theta }{2}+2\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}}{{\mathrm{cos}}^{2}\frac{\theta }{2}+{\mathrm{sin}}^{2}\frac{\theta }{2}-2\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{y+1}{1-y}=\sqrt{\frac{{\left(\mathrm{co}s\frac{\theta }{2}+\mathrm{sin}\frac{\theta }{2}\right)}^{2}}{{\left(\mathrm{co}s\frac{\theta }{2}-\mathrm{sin}\frac{\theta }{2}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{y+1}{1-y}=\frac{\left(\mathrm{co}s\frac{\theta }{2}+\mathrm{sin}\frac{\theta }{2}\right)}{\left(\mathrm{co}s\frac{\theta }{2}-\mathrm{sin}\frac{\theta }{2}\right)}\left[\because 0<\theta <\frac{\mathrm{\pi }}{2}⇒0<\frac{\theta }{2}<\frac{\mathrm{\pi }}{4},0\mathrm{to}\frac{\mathrm{\pi }}{4}\mathrm{cos}\theta \mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{sin}\theta \right]\phantom{\rule{0ex}{0ex}}⇒\frac{y+1}{1-y}=\frac{\frac{\mathrm{co}s\frac{\theta }{2}}{\mathrm{co}s\frac{\theta }{2}}+\frac{\mathrm{sin}\frac{\theta }{2}}{\mathrm{co}s\frac{\theta }{2}}}{\frac{\mathrm{co}s\frac{\theta }{2}}{\mathrm{co}s\frac{\theta }{2}}-\frac{\mathrm{sin}\frac{\theta }{2}}{\mathrm{co}s\frac{\theta }{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+y}{1-y}=\frac{1+\mathrm{tan}\frac{\theta }{2}}{1-\mathrm{tan}\frac{\theta }{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Comparing}\mathrm{both}\mathrm{the}\mathrm{sides}:\phantom{\rule{0ex}{0ex}}y=\mathrm{tan}\frac{\theta }{2}$

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