$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
$\sin \frac{x}{2}$
$\tan \frac{x}{2}$
$\sec \frac{x}{2}$
${\rm{cosec}}\frac{x}{2}$
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
The value of $\cos 15^\circ - \sin 15^\circ $ is equal to
Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then
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 :-