If $\tan A = \frac{1 - \cos B}{\sin B},$ find $\tan 2A$ in terms of $\tan B$.

  • A
    $\tan 2A = \tan B$
  • B
    $\tan 2A = \tan^2 B$
  • C
    $\tan 2A = \tan^2 B + 2\tan B$
  • D
    None of the above

Explore More

Similar Questions

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $

If $\tan \frac{\theta}{2} = t$,then $\frac{1 - t^2}{1 + t^2}$ is equal to

If $k = \sin \frac{\pi}{18} \cdot \sin \frac{5\pi}{18} \cdot \sin \frac{7\pi}{18}$,then the numerical value of $k$ is

Evaluate: $\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5}$

If $\sin^4 \theta \cos^2 \theta = \sum_{n=0}^{\infty} a_{2n} \cos 2n \theta$,then the least $n$ for which $a_{2n} = 0$ is

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