$\tan \frac{A}{2}$ is equal to

  • A

    $ \pm \sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} $

  • B

    $ \pm \sqrt {\frac{{1 + \sin A}}{{1 - \sin A}}} $

  • C

    $ \pm \sqrt {\frac{{1 - \cos A}}{{1 + \cos A}}} $

  • D

    $ \pm \sqrt {\frac{{1 + \cos A}}{{1 - \cos A}}} $

Similar Questions

If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

Value of $\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ is equal to

Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

If $\sin \alpha = \frac{{ - 3}}{5},$ where $\pi < \alpha < \frac{{3\pi }}{2},$ then $\cos \frac{1}{2}\alpha = $