If $x=\sqrt{a^{sin^{-1}t}},y = \sqrt{a^{cos^{-1}t}},$ then $\frac{dy}{dx}$ is :

$-\frac{y}{x}$

The correct answer is Option (2) → $-\frac{y}{x}$

$x=\sqrt{a^{sin^{-1}t}},y = \sqrt{a^{cos^{-1}t}}$

$\log x=\frac{sin^{-1}t}{2}\log a$, $\log y=\frac{cos^{-1}t}{2}\log a$

differentiating wrt (t)

$\frac{1}{x}\frac{dx}{dt}=\frac{1}{2\sqrt{1-t^2}\log a}$  ...(1)

$\frac{1}{y}\frac{dy}{dt}=\frac{-1}{2\sqrt{1-t^2}\log a}$

so $\frac{1}{x}\frac{dx}{dt}=\frac{-1}{y}\frac{dy}{dt}$

$⇒\frac{dy}{dx}=-\frac{y}{x}$