Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

If the inverse trigonometric functions take principal values,then $\cos ^{-1}\left(\frac{3}{10} \cos \left(\tan ^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5} \sin \left(\tan ^{-1}\left(\frac{4}{3}\right)\right)\right)$ is equal to

${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

If $f(x) = \frac{\sqrt{\operatorname{Cos}^{-1} x}}{\sqrt{2(1-x)}}$ for $x < 1$,then $\lim_{x \rightarrow 1^{-}} f(x) =$

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $0 < x < 1$ and $0 < y < \frac{\pi}{2}$,then $\frac{dy}{dx} = $

$\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=$ . . . . . . .