In a triangle $\tan A + \tan B + \tan C = 6$ and $\tan A\tan B = 2,$ then the values of $\tan A,\,\,\tan B$ and $\tan C$ are
$1, 2, 3$
$2,1,3$
$(a)$ and $(b)$ both
None of these
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Show that
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
If $\alpha ,\,\beta ,\,\gamma \in \,\left( {0,\,\frac{\pi }{2}} \right)$, then $\frac{{\sin \,(\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }}$ is
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then