If $ heta$ lies in the second quadrant, then the value of $\sqrt{\frac{1 - \sin heta}{1 + \sin heta}} + \sqrt{\frac{1 + \sin heta}{1 - sin heta}}$ is equal to

Question

If $	heta$ lies in the second quadrant, then the value of $\sqrt{\frac{1 - \sin	heta}{1 + \sin	heta}} + \sqrt{\frac{1 + \sin	heta}{1 - sin	heta}}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\sqrt{\frac{1 - \sin	heta}{1 + sin\ 	heta}} + \sqrt{\frac{1 + \sin	heta}{1 - \sin	heta} =}\frac{1 - \sin{	heta + 1 + \sin	heta}}{\sqrt{1 - \sin^{2}	heta}}$
$= \frac{2}{\sqrt{\cos^{2}	heta}} = \frac{2}{|\cos	heta|}$
$= - \frac{2}{\cos	heta} = - 2\sec{	heta\left( \because\frac{\pi}{2} < 	heta < \pi ight)}$

Q. If $\theta$ lies in the second quadrant, then the value of $\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} + \sqrt{\frac{1 + \sin\theta}{1 - sin\theta}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?