Question

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

• A. $\sqrt{\frac{1-y^2}{1-x^2}}$
• B. $\sqrt{\frac{y}{x}}$
• C. $\frac{\pi }{2}$
• D. $-\frac{y}{x}$

Answer 1

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

$x=a^{{\sin }^{-1}t}\Rightarrow {\log }_{a}x={\sin }^{-1}t$
$y=a^{{\cos }^{-1}t}\Rightarrow {\log }_{a}y={\cos }^{-1}t$
$\therefore {\log }_{a}x+{\log }_{a}y={\sin }^{-1}t+{\cos }^{-1}t=\frac{\pi }{2}$
$\Rightarrow {\log }_{a}xy=\frac{\pi }{2}\Rightarrow xy=a^{\frac{\pi }{2}}$
$\Rightarrow x\frac{dy}{dx}+y=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$