If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$$ = \tan \alpha \tan \beta \tan \gamma $, then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $
$\cot \alpha \cot \beta \cot \gamma $
$\tan \alpha \tan \beta \tan \gamma $
$\cot \alpha + \cot \beta + \cot \gamma $
$\tan \alpha + \tan \beta + \tan \gamma $
If $p = \frac{{2\sin \,\theta }}{{1 + \cos \theta + \sin \theta }}$, and $q = \frac{{\cos \theta }}{{1 + \sin \theta }},$ then
If $\sin {\theta _1} + \sin {\theta _2} + \sin {\theta _3} = 3,$ then $\cos {\theta _1} + \cos {\theta _2} + \cos {\theta _3} = $
Let $A, B$ and $C$ are the angles of a plain triangle and $\tan \frac{A}{2} = \frac{1}{3},\,\,\tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to
Prove that: $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
If $5\tan \theta = 4,$ then $\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = $