$\frac{\sqrt{2}-\sin \alpha-\cos \alpha}{\sin \alpha-\cos \alpha}=$

  • A
    $\sec \left(\frac{\alpha}{2}-\frac{\pi}{8}\right)$
  • B
    $\cos \left(\frac{\pi}{8}-\frac{\alpha}{2}\right)$
  • C
    $\tan \left(\frac{\alpha}{2}-\frac{\pi}{8}\right)$
  • D
    $\cot \left(\frac{\alpha}{2}-\frac{\pi}{2}\right)$

Explore More

Similar Questions

$\operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\operatorname{coth}^{-1} 3\right)=$

The value of $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}$ is

Difficult
View Solution

Let $S = \{x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) : 9^{1-\tan^2 x} + 9^{\tan^2 x} = 10\}$ and $\beta = \sum_{x \in S} \tan^2\left(\frac{x}{3}\right)$,then $\frac{1}{6}(\beta - 14)^2$ is equal to

If $\cosh \beta = \sec \alpha \cos \theta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin \theta$,then $\sinh^2 \beta =$

If $(\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)$,then each side is equal to

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