Question

If $\pi \le \theta \le \frac{3\pi }{2}$ then $\sqrt{}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }+\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}$ is equal to :

• A. 2 $co\sec \theta$
• B. -2 $co\sec \theta$
• C. 2 $\sec \theta$
• D. $-\sec \theta$

Answer 1

The correct option is B -2 $co\sec \theta$

Given expression
$=\frac{(1+\cos \theta )+(1-\cos \theta )}{\sqrt{1-{\cos }^2\theta }}$
$=\frac{2}{\sin \theta }$
$\because \theta$ in the 3rd quardrant,so $\sin \theta$ is negative
$\therefore$ Given expression $=-\frac{2}{\sin \theta }=-2co\sec \theta$