The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

The number of all the values of $x$ for which the function $f(x)=\sin x+\frac{1-\tan ^2 x}{1+\tan ^2 x}$ attains its maximum value on $[0, 2\pi]$ is

If $\alpha \in \left( 0, \frac{\pi}{2} \right)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$ is always greater than or equal to

If $\frac{\sin^4 A}{a} + \frac{\cos^4 A}{b} = \frac{1}{a + b},$ then the value of $\frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3}$ is equal to

The value of $k$,for which $(\cos x + \sin x)^2 + k \sin x \cos x - 1 = 0$ is an identity,is

Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta)=4(\sin^{4}(\frac{7\pi}{2}-\theta)+\sin^{4}(11\pi+\theta)) - 2(\sin^{6}(\frac{3\pi}{2}-\theta)+\sin^{6}(9\pi-\theta))$, $\theta \in R$. Then $\alpha+2\beta$ is equal to: