The expression $\frac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$ is equal to
$\cos 2x$
$2\cos x$
${\cos ^2}x$
$1 + \cos x$
If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \,\left( {\frac{{x + y}}{2}} \right) = $
If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$
Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $
If $A + B + C = {180^o},$ then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ will be
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $