$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$0$
$1$
$2$
$4$
If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ then $\tan (\alpha+2 \beta)$ is equal to
Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
If $A + B + C = \pi ,$ then ${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + $${\tan ^2}\frac{C}{2}$ is always
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$