If cos ^{-1}left(frac{x^2-y^2}{x^2+y^2}right) | vedclass

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(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$.Taking $\cos$ on both sides,we get $\frac{x^2-y^2}{x^2+y^2} = \cos(\sin^{-1}

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$y / x$

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(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$.Taking $\cos$ on both sides,we get $\frac{x^2-y^2}{x^2+y^2} = \cos(\sin^{-1} a)$.Let $\cos(\sin^{-1} a) = k$,where $k$ is a constant.So,$\frac{x^2-y^2}{x^2+y^2} = k$.$x^2 - y^2 = k(x^2 + y^2) \Rightarrow x^2(1-k) = y^2(1+k)$.$y^2 = x^2 \left(\frac{1-k}{1+k}\right)$.Let $C = \sqrt{\frac{1-k}{1+k}}$,which is a constant.Then $y^2 = C^2 x^2 \Rightarrow y = Cx$ (assuming $y, x > 0$ for simplicity).Differentiating with respect to $x$,we get $\frac{dy}{dx} = C$.Since $y = Cx$,we have $C = \frac{y}{x}$.Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

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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