Inverse Trigonometric Functions

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