यदि $A + B + C = \frac{{3\pi }}{2},$ तब $\cos 2A + \cos 2B + \cos 2C = $
यदि $A + B + C = \frac{{3\pi }}{2},$ तब $\cos 2A + \cos 2B + \cos 2C = $
- A
$1 - 4\cos A\,\cos B\,\cos C$
- B
$4\sin A\,\,\sin B\,\,\sin C$
- C
$1 + 2\cos A\,\cos B\,\cos C$
- D
$1 - 4\sin A\,\,\sin B\,\,\sin C$
Similar Questions
$\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)}} = $
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
दिखाइए
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
दिखाइए
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
यदि $A + B + C = {180^o},$ तो $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
यदि $A + B + C = {180^o},$ तो $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $