The value of $\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}}$ is equal to

$2\cos x - \cos 3x - \cos 5x = $

$\frac{{\cos A}}{{1 - \sin A}} = $

Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$

The expression $\frac{{{{\tan }^2}20^\circ  - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to

If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $