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
$-cot \frac{x}{2}$
$cot \frac{x}{2}$
$ tan \frac{x}{2}$
$-tan \frac{x}{2}$
The expression $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to
$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
If $\sin \alpha = \frac{{336}}{{625}}$ and $450^\circ < \alpha < 540^\circ ,$ then $\sin \left( {\frac{\alpha }{4}} \right) = $
$\sqrt {2 + \sqrt {2 + 2\cos 4\theta } } = $
If $A + B + C = {180^o},$ then the value of $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B)$ will be