If $x=\tan ^{-1}\left\{\frac{\sqrt{1+t^2}-1}{t}\right\}, y=\cos ^{-1}\left\{\frac{1-t^2}{1+t^2}\right\}, \quad$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

$$ \cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty= $$

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$, then the value of $\sin x$ is