The solution set of the inequation $${\cos ^{ - 1}}x < {\sin ^{ - 1}}x$$ is

If $\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 r^2}\right)=a$, then $\tan a$ is equal to

The true set of values of ' $K$ ' for which $\sin ^{-1}\left(\frac{1}{1+\sin ^2 x}\right)=\frac{K \pi}{6}$ may have a solution is

If for two real numbers $\mathrm{a}, \mathrm{b}$ with $|\mathrm{a}| \leq 1$ and $|\mathrm{b}| \leq 1$,

$\frac{1}{3}+\frac{\sin ^{-1} a+\sin ^{-1} b}{4}+\frac{\left(\sin ^{-1} a+\sin ^{-1} b\right)^2}{16}+\frac{\left(\sin ^{-1} a+\sin ^{-1} b\right)^3}{64}+\cdots=\frac{2(8-3 \pi)}{3(16+3 \pi)}, \quad$ then the value of $\sin ^{-1}\left(a \sqrt{1-b^2}+b \sqrt{1-a^2}\right)$ is