Question

Find $m$ if $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}=\frac{-m}{\cos \theta },\frac{\pi }{2}<\theta <\pi$.

Answer 1

$\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$

$\Rightarrow \sqrt{\frac{(1-\sin \theta {)}^2}{1-{\sin }^2\theta }}+\sqrt{\frac{(1+\sin \theta {)}^2}{1-{\sin }^2\theta }}$

$\because {\sin }^{2}A+{\cos }^{2}A=1$

$\because (a+b)(a-b)=a^2-b^2$

$\Rightarrow \left|\frac{1-\sin \theta }{\cos \theta }\right|+\left|\frac{1+\sin \theta }{\cos \theta }\right|$

$\because \theta ϵ(\pi /2,\pi )$

So, $\sin \theta >0$ & $\cos \theta <0$

$\Rightarrow \frac{1-\sin \theta }{-\cos \theta }+\frac{1+\sin \theta }{-\cos \theta }$

$\Rightarrow \frac{-2}{\cos \theta }=\frac{-m}{\cos \theta }$

$\Rightarrow m=-2$