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

For the least possible value of $n \in Z$,the solution $(x, y)$ of the equations $\cos ^{-1} x + (\sin ^{-1} y)^2 = \frac{n \pi^2}{4}$ and $(\cos ^{-1} x)(\sin ^{-1} y)^2 = \frac{\pi^4}{16}$ is

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$,then $8x^2 + 1 =$

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then $x(y+z)+y(z+x)+z(x+y)$ equals to

If $\sinh ^{-1}(-\sqrt{3})+\cosh ^{-1}(2)=K$,then $\cosh K=$

Define $f: R \rightarrow R$ by $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$. Then $\lim_{x \rightarrow \infty} (f \circ f)(x)$ is equal to