For a >0, t ∈(0, (π)/2), let x=√(a^sin^-1 t) and y=√(a^cos^-1 t). Then, 1+((d y)/(d x))^2 equals:

For $\displaystyle \mathrm{a}>0, \mathrm{t} \in\left(0, \frac{\pi}{2}\right)$, let $\displaystyle x=\sqrt{\mathrm{a}^{\sin ^{-1} \mathrm{t}}}$ and $\displaystyle y=\sqrt{\mathrm{a}^{\cos ^{-1} \mathrm{t}}}$. Then, $\displaystyle 1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2$ equals :