The derivative of ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ with respect to ${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ is

$\pi + \left(\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65}\right)$ is equal to

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

If $\alpha, \beta$ 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 $\tan^{-1} (x^2 + 3|x|-4) = \tan^{-1} (4\pi + \sin^{-1}(\sin 14))$,then the value of $\cos^{-1}(\cos 3|x|)$ is equal to

The derivative of the function $f(x) = \cos^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x - 3\sin x) \right\} + \sin^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x + 3\sin x) \right\}$ with respect to $x$ at $x = \frac{3}{4}$ is: