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

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(A) Given, $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ $\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos k$ Let $\cos k = C$ (a constant). Then, $x^2 -

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$\frac{y}{x}$

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(A) Given, $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ $\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos k$ Let $\cos k = C$ (a constant). Then, $x^2 - y^2 = C(x^2 + y^2)$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(C(x^2 + y^2))$ $2x - 2y \frac{dy}{dx} = C(2x + 2y \frac{dy}{dx})$ $x - y \frac{dy}{dx} = C(x + y \frac{dy}{dx})$ $x - Cx = Cy \frac{dy}{dx} + y \frac{dy}{dx}$ $x(1 - C) = y \frac{dy}{dx}(C + 1)$ $\frac{dy}{dx} = \frac{x(1 - C)}{y(1 + C)}$ Substituting $C = \frac{x^2-y^2}{x^2+y^2}$ back: $\frac{dy}{dx} = \frac{x(1 - \frac{x^2-y^2}{x^2+y^2})}{y(1 + \frac{x^2-y^2}{x^2+y^2})} = \frac{x(\frac{x^2+y^2-x^2+y^2}{x^2+y^2})}{y(\frac{x^2+y^2+x^2-y^2}{x^2+y^2})} = \frac{x(2y^2)}{y(2x^2)} = \frac{xy^2}{yx^2} = \frac{y}{x}$

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

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