The value of $\cot {70^o} + 4\cos {70^o}$ is
$\frac{1}{{\sqrt 3 }}$
$\sqrt 3 $
$2\sqrt 3 $
$\frac{1}{2}$
$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
If $\alpha $ and $\beta $ are solutions of $sin^2\,x + a\, sin\, x + b = 0$ as well that of $cos^2\,x + c\, cos\, x + d = 0$ , then $sin\,(\alpha + \beta )$ is equal to
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $