If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$ then $\frac{d y}{d x}$ is equal to

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Aicharya · Accepted Answer

Given, $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1} a$ $ \Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos \left(\sin ^{-1} a\right)=c $ On applying componendo and dividendo law, we get $ \frac{2 x^2}{2 y^2}=\frac{c+1}{1-c} \Rightarrow \frac{y^2}{x^2}=\frac{1-c}{1+c} $ On differentiating both sides w.r.t ' $x$ ', we get $ 2 x^2 y \frac{d y}{d x}-2 y^2 x=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x} $

If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$ then $\frac{d y}{d x}$ is equal to

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