Question

IF $-\pi <\theta <-\frac{\pi }{2}$, then $|\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}|$ is equal to

• A. $2\sec \theta$
• B. $-2\sec \theta$
• C. $2{sec}^2\theta$
• D. $2\sec \frac{\theta }{2}$

Answer 1

The correct option is A $2\sec \theta$

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

$\left|\frac{1-\sin \theta +1+\sin \theta }{1-{\sin }^2\theta }\right|=\left|\frac{2}{\sqrt{1-{\sin }^2\theta }}\right|=\left|\frac{2}{\cos \theta }\right|$

$=|2|\sec \theta ||$

$=2\sec \theta$