$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
$\frac{1}{2}\tan \theta $
$\frac{1}{2}\cot \theta $
$\tan \theta $
$\cot \theta $
If $\tan \theta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }},$ then $\sin \alpha + \cos \alpha $ and $\sin \alpha - \cos \alpha $ must be equal to
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then
If $\tan x = \frac{b}{a},$ then $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
If $3\cos \theta + 4\sin \theta = 5$ then $3\sin \theta - 4\cos \theta $ is