Question

If $0<\theta <\frac{\pi }{2}$, and if $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$, then $y$ is equal to

• A. $\cot \frac{\theta }{2}$
• B. $\tan \frac{\theta }{2}$
• C. $\cot \frac{\theta }{2}+\tan \frac{\theta }{2}$
• D. $\cot \frac{\theta }{2}-\tan \frac{\theta }{2}$

Answer 1

The correct option is B $\tan \frac{\theta }{2}$

$\left(\frac{1+y}{1-y}\right)=\sqrt{\frac{1+sin\theta }{1-sin\theta }}$

$\left(\frac{1+y}{1-y}\right)=\sqrt{\frac{(sin\theta /2+cos\theta /2{)}^2}{(sin\theta /2-cos\theta /2{)}^2}}$

$\left(\frac{1+y}{1-y}\right)=\frac{|sin\theta /2+cos\theta /2|}{|cos\theta /2-sin\theta /2|}$

since $\theta /2E[0,\pi /4]$

$\left(cos\theta /2\right)>\left(sin\theta /2\right)$

$\Rightarrow \left(\frac{1+y}{1-y}\right)=\left(\frac{sin\theta /2+cos\theta /2}{cos\theta /2-sin\theta /2}\right)$

Applying compound dividend

$\Rightarrow \left(\frac{2}{2y}\right)=\frac{2cos\theta /2}{2sin\theta /2}$

$\Rightarrow y=tan\left(\frac{\theta }{2}\right)$