If $x \in (0, \frac{\pi}{4})$,then the expression $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ cannot take which of the following values?

If $\alpha+\beta+\gamma=2 \theta$,then $\cos \theta+\cos (\theta-\alpha)+\cos (\theta-\beta)+\cos (\theta-\gamma)$ is equal to

The extremum values of the function $f(x) = \frac{1}{\sin x + 4} - \frac{1}{\cos x - 4}$ for $x \in R$ are:

Let $f(\theta) = (1 + \sin^2 \theta)(2 - \sin^2 \theta)$. Then, for all values of $\theta$:

$a, b, c$ are the sides of a scalene triangle $ABC$. If angles $\alpha, \beta, \gamma$ lie between $0$ and $\pi$ such that $\cos \alpha = \frac{a}{b+c}, \cos \beta = \frac{b}{c+a}$ and $\cos \gamma = \frac{c}{a+b}$,then $\tan^2 \frac{\alpha}{2} + \tan^2 \frac{\beta}{2} + \tan^2 \frac{\gamma}{2} =$

$\cos 2\theta + 2\cos \theta$ is always