જો $\tan \beta = \cos \theta \tan \alpha ,$ તો ${\tan ^2}\frac{\theta }{2} = $
જો $\tan \beta = \cos \theta \tan \alpha ,$ તો ${\tan ^2}\frac{\theta }{2} = $
- A
$\frac{{\sin (\alpha + \beta )}}{{\sin (\alpha - \beta )}}$
- B
$\frac{{\cos (\alpha - \beta )}}{{\cos (\alpha + \beta )}}$
- C
$\frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$
- D
$\frac{{\cos (\alpha + \beta )}}{{\cos (\alpha - \beta )}}$
Similar Questions
જો $A + B + C = {180^o},$ તો $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
જો $A + B + C = {180^o},$ તો $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
$\tan 75^\circ - \cot 75^\circ = $
$\tan 75^\circ - \cot 75^\circ = $
સાબિત કરો કે : $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
સાબિત કરો કે : $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$