If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$, then $\frac{d y}{d x}$ is equal to
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If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$, then $\frac{d y}{d x}$ is equal to

(A) $\frac{x}{y}$

(B) $-\frac{x}{y}$

(C) $\frac{y}{x}$

(D) $-\frac{y}{x}$

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Best answer

SOLUTION —

$\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \Rightarrow \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1} y$

$\Rightarrow \quad \sin ^{-1} x=\cos ^{-1} y \Rightarrow y=\sqrt{1-x^2}$

$\Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}}(-2 x)=-\frac{x}{y}$

So, The correct option will be (B).

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