$\sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} \right)} \right)} $ is equal to-

Let $\alpha = 3 \sin^{-1}(\frac{6}{11})$ and $\beta = 3 \cos^{-1}(\frac{4}{9})$, where inverse trigonometric functions take only the principal values. Given below are two statements:
Statement $I$: $\cos(\alpha + \beta) > 0$.
Statement $II$: $\cos(\alpha) < 0$.
In the light of the above statements, choose the correct answer from the options given below:

If the quadratic equation $4^{\sec^2 \alpha} x^2 + 2x + (\beta^2 - \beta + \frac{1}{2}) = 0$ has real roots,then the value of $\cos^2 \alpha + \cos^{-1} \beta$ is

If $p$ and $q$ are roots of $6x^2 + 10x + 1 = 0$,then the value of $[\tan^{-1} p + \tan^{-1} q]$ is: {where $[x]$ denotes the greatest integer less than or equal to $x$}

$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to