The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$ is

Let $x=a \sin ^\alpha \theta \cos ^{\alpha+1} \theta$ and $y=a \sin ^{\alpha+1} \theta \cos ^\alpha \theta$,where $\theta \neq \frac{n \pi}{2}$. If $\frac{(x^2+y^2)^m}{(xy)^n}$ is independent of $\theta$,then the relation between $\alpha, m$ and $n$ is:

If $\operatorname{cosec} \theta + \cot \theta = 5$,then $\sin \theta = $

$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} = $

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

$1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ = $