If $\sin x + \sin^2 x = 1$,then the value of the expression $\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x - 1$ is equal to

If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma ) = \tan \alpha \tan \beta \tan \gamma $,then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma ) = $

$\cos 2(\theta + \phi) - 4\cos (\theta + \phi)\sin \theta \sin \phi + 2\sin^2 \phi = $

If $\cos \theta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$,then one of the values of $\tan \frac{\theta}{2}$ is

Let $\alpha$ and $\beta$ be real numbers such that $-\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4}$. If $\sin (\alpha+\beta) = \frac{1}{3}$ and $\cos (\alpha-\beta) = \frac{2}{3}$,then the greatest integer less than or equal to $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$ is:

Let $|\cos \theta \cos (60^{\circ}-\theta) \cos (60^{\circ}+\theta)| \leq \frac{1}{8}$, where $\theta \in [0, 2\pi]$. Then, the sum of all $\theta \in [0, 2\pi]$ where $\cos 3\theta$ attains its maximum value is: (in $\pi$)