Q. If$\ \frac{3\pi}{4} < \alpha < \pi,\ then\ \sqrt{cosec^{2}\alpha + 2\cot\alpha}\$is equal to

A) $1 + \cot\alpha$
B) $1 - \cot\alpha$
C) $- 1 - \cot\alpha$
D) $- 1 + \cot\ \alpha$
Explanation

(c)
We have, $\sqrt{cosec^{2}\alpha + 2\cot\alpha}$
$= \sqrt{1 + \cot^{2}{\alpha + 2\cot\alpha}} = \left| 1 + \cot\alpha \right|$
But $\frac{3\pi}{4} < \alpha < \pi$
$\Rightarrow \cot{\alpha < - 1\ \Rightarrow 1 + \cot{\alpha < 0}}$
Hence, $\left| 1 + \cot\alpha \right| = \ - (1 + \cot{\alpha)}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.