Question

If $x=\sqrt{a^{{\sin }^{-1}t}}$, $y=\sqrt{a^{{\cos }^{-1}t}}$, $a>0$ and $-1<t<1$, $\frac{dy}{dx}=$

• A. $\frac{x}{y}$
• B. $-\frac{x}{y}$
• C. $-\frac{y}{x}$
• D. $\frac{y}{x}$

Answer 1

The correct option is C $-\frac{y}{x}$

We have,
$x=\sqrt{a^{{\sin }^{-1}t}}$, $y=\sqrt{a^{{\cos }^{-1}t}}$
Multiplying $x$ and $y$
$x.y=\sqrt{a^{{\sin }^{-1}t+{\cos }^{-1}t}}$ ........................Since, ${\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}$
$=\sqrt{a^{\pi /2}}$
Differentiating w.r.t. $x$, we get
$x\frac{dy}{dx}+y=0$
$\frac{dy}{dx}=\frac{-y}{x}$