If $y=\tanh ^{-1} \sqrt{\frac{1-x}{1+x}}$, then $\frac{d y}{d x}=$

$$ \tan ^{-1} \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}+\tan ^{-1} \frac{1}{\sqrt{5}}= $$

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

If $0 < x < \frac{1}{2}$ and $\alpha=\sin ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3-3 x^2}}{2}\right)$, then $\tan \alpha+\cot \alpha$ is equal to