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

Let $g(x)=a x+b$, where $a<0$ and $g$ is defined from $[1,3]$ onto $[0,2]$. Then the value of $\cot \left(\cos ^{-1}(|\sin x|+|\cos x|)+\right. \left.\sin ^{-1}(-|\cos x|-|\sin x|)\right)$ is equal to

If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ is equal to