If $\sin 5 \theta = \cos 20^{\circ}$ where $0^{\circ} < \theta < 90^{\circ}$,then the value of $\theta$ is (in degrees):

$\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}} = $

Let $\alpha, \beta$ be such that $\pi < (\alpha - \beta) < 3\pi$. If $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$,then the value of $\cos \frac{\alpha - \beta}{2}$ is

Let $A, B$ and $C$ be the angles of a plane triangle and $\tan \frac{A}{2} = \frac{1}{3}, \tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$,where $0 \le \alpha, \beta \le \frac{\pi}{4}$. Then $\tan 2\alpha = $

$\frac{\tan ^{2} \theta}{\sec \theta+1}-\sec \theta$ is equal to