The derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right)$ with respect to $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$ is...

If $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$,then $x=$ . . . . . . .

For $k \in R$,let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k$,where $0 < |x| < \frac{1}{\sqrt{2}}$,be $\alpha$ and $\beta$,where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2}- bx -5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$,then $\frac{b}{k^{2}}$ is equal to $......$

If $y = \sum_{k=1}^{6} k \cos^{-1} \left\{ \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right\}$,then $\frac{dy}{dx}$ at $x = 0$ is

For $a>0$, if $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$, then $\cot \left[\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]=$

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :