If $\frac{5\pi}{2} < x < 3\pi$,then the value of the expression $\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}$ is

  • A
    $-\cot \frac{x}{2}$
  • B
    $\cot \frac{x}{2}$
  • C
    $\tan \frac{x}{2}$
  • D
    $-\tan \frac{x}{2}$

Explore More

Similar Questions

The maximum value of $3 \cos \theta + 5 \sin \left( \theta - \frac{\pi}{6} \right)$ for any real value of $\theta$ is

Difficult
View Solution

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

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$ then $\cos 2\beta$ is equal to

$\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}} = $

If $7 \sin^{2} \theta + 3 \cos^{2} \theta = 4$,then the value of $\tan \theta$ is (where $\theta$ is acute).

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