If $A, B, C$ are angles of a triangle, then $\sin 2A + \sin 2B - \sin 2C$ is equal to
$4\sin A\,\,\cos B\,\,\cos C$
$4\cos A$
$4\sin A\,\cos A$
$4\cos A\,\cos B\,\sin C$
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
If $\tan A = \frac{1}{2},$ then $\tan 3A = $
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is
If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha $and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha $, then $\theta$ is equal to