$\tan \frac{A}{2} = . . .$
$\tan \frac{A}{2} = . . .$
- 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
$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ =
$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ =
જો ${\tan ^2}\theta = 2{\tan ^2}\phi + 1,$ તો $\cos 2\theta + {\sin ^2}\phi = . . .$
જો ${\tan ^2}\theta = 2{\tan ^2}\phi + 1,$ તો $\cos 2\theta + {\sin ^2}\phi = . . .$
સાબિત કરો કે : $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
સાબિત કરો કે : $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
જો $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta ) = . . . .$
જો $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ તો ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta ) = . . . .$