$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
- A
$0$
- B
$\frac{1}{2}$
- C
$\frac{1}{4}$
- D
$ - \frac{1}{8}$
Similar Questions
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}=...............$
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}=...............$
- [JEE MAIN 2023]
જો $90^\circ < A < 180^\circ $ અને $\sin A = \frac{4}{5},$ તો $\tan \frac{A}{2} = . . .$
જો $90^\circ < A < 180^\circ $ અને $\sin A = \frac{4}{5},$ તો $\tan \frac{A}{2} = . . .$
$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}} =$
$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}} =$
- [IIT 1985]
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =
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}} = $