If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

  • A
    $\frac{2}{\sin \alpha}$
  • B
    $-\frac{2}{\sin \alpha}$
  • C
    $\frac{1}{\sin \alpha}$
  • D
    $-\frac{1}{\sin \alpha}$

Explore More

Similar Questions

The minimum value of $(8 \sec^2 \theta + 2 \cos^2 \theta)$ is equal to :-

If $\tan 80^{\circ} = a$ and $\tan 47^{\circ} = b$,then $\tan 37^{\circ}$ is equal to -

The value of $\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}$ is

If $\sqrt x + \frac{1}{{\sqrt x }} = 2\cos \theta ,$ then ${x^6} + {x^{ - 6}} = $

$\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo