$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = $

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1, 1]$ such that $\cos ^{-1} x - \sin ^{-1} y = \alpha$,where $-\frac{\pi}{2} \leq \alpha \leq \pi$. Then,the minimum value of $x^2 + y^2 + 2xy \sin \alpha$ is

If $x, y, z$ are in arithmetic progression and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in arithmetic progression,then:

$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to..........

If $\alpha, \beta$ are the solutions of the equation $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$ and $\alpha > \beta$,then $3\alpha + 4\beta =$

If ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi ,$ then the value of $a\sqrt {1 - {a^2}} + b\sqrt {1 - {b^2}} + c\sqrt {1 - {c^2}}$ is