જો ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ તો $K =$
જો ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ તો $K =$
- A
$\frac{4}{3}$
- B
$\frac{3}{4}$
- C
$\frac{1}{2}$
- D
$2$
Similar Questions
$\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} = $
$4 \,\,sin5^o \,\,sin55^o \,\,sin65^o$ =
$4 \,\,sin5^o \,\,sin55^o \,\,sin65^o$ =
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
- [IIT 1986]
સાબિત કરો કે, $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
સાબિત કરો કે, $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $