If $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$,then the derivative of $f(x)$ with respect to $g(x)$ is

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

If $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ and $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$,then $\frac{dy}{dx} = $

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$ is: (in $\pi^2$)

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

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