Question

Given $\frac{\pi }{2}<\alpha <\pi$, then the expression $\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }}=$

• A. $\frac{1}{\cos \alpha }$
• B. $-\frac{2}{\cos \alpha }$
• C. $\frac{2}{\cos \alpha }$
• D. None of these

Answer 1

Answer: (B)

$-\frac{2}{\cos \alpha }$

Given, $\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }}=\frac{(1-\sin \alpha )+(1+\sin \alpha )}{\sqrt{(1+\sin \alpha )(1-\sin \alpha )}}$

$=\frac{2}{\sqrt{{\left({\cos }^2\alpha \right)}^2}}=\frac{2}{\left|\cos \alpha \right|}=-\frac{2}{\cos \alpha }$

$\because \frac{\pi }{2}<\alpha <\pi \Rightarrow \cos \alpha <0\Rightarrow \left|\cos \alpha \right|=-\cos \alpha$