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

The value of $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ $ is

If $\frac{{5\pi }}{2} < x < 3\pi $, then the value of the expression $\frac{{\sqrt {1 - \sin x}  + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x}  - \sqrt {1 + \sin x} }}$ is

If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-

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

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 $\tan 2\alpha =$