If tan y = frac{x sin alpha}{1-x cos alpha} a | vedclass

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(C) Given $	an y = \frac{x \sin \alpha}{1-x \cos \alpha}$.Taking the derivative with respect to $x$ on both sides:$\frac{d}{dx}(	an y) = \frac{d}{dx

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(C) Given $	an y = \frac{x \sin \alpha}{1-x \cos \alpha}$.Taking the derivative with respect to $x$ on both sides:$\frac{d}{dx}(	an y) = \frac{d}{dx}\left(\frac{x \sin \alpha}{1-x \cos \alpha}ight)$$\sec^2 y \cdot \frac{dy}{dx} = \frac{(1-x \cos \alpha)(\sin \alpha) - (x \sin \alpha)(-\cos \alpha)}{(1-x \cos \alpha)^2}$$\sec^2 y \cdot \frac{dy}{dx} = \frac{\sin \alpha - x \sin \alpha \cos \alpha + x \sin \alpha \cos \alpha}{(1-x \cos \alpha)^2} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$Since $	an y = \frac{x \sin \alpha}{1-x \cos \alpha}$,we have $\sec^2 y = 1 + 	an^2 y = 1 + \frac{x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{(1-x \cos \alpha)^2 + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2 \cos^2 \alpha + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2}$.Substituting this back:$\frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2} \cdot \frac{dy}{dx} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$$\frac{dy}{dx} = \frac{\sin \alpha}{x^2 - 2x \cos \alpha + 1}$.Comparing this with $\frac{dy}{dx} = \frac{m}{x^2+2nx+1}$,we get $m = \sin \alpha$ and $n = -\cos \alpha$.Therefore,$m^2 + n^2 = \sin^2 \alpha + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1$.

If $\tan y = \frac{x \sin \alpha}{1-x \cos \alpha}$ and $\frac{dy}{dx} = \frac{m}{x^2+2nx+1}$,then $m^2+n^2$ is

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