Question

If $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$, then the value of $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$ is equal to

• A. $-\sec \theta +\tan \theta$
• B. $\sec \theta +\tan \theta$
• C. $\sec \theta -\tan \theta$
• D. None of these

Answer 1

The correct option is A $-\sec \theta +\tan \theta$

The value of $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$ is

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

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

$=\frac{1-\sin \theta }{-\cos \theta }(\because \frac{\pi }{2}<\theta <\frac{3\pi }{2})$

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

$=-\sec \theta +\tan \theta$.