If \(\tan y=\frac{x \sin \alpha}{1-x \cos \alpha}\) and \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{m}}{x^2+2 \mathrm{n} x+1}\), then \(m^2+n^2\) is

\(\begin{aligned} & \tan y=\frac{x \sin \alpha}{1-x \cos \alpha} \\ & \therefore \quad y=\tan ^{-1}\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right) \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right)^2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right) \\ & =\frac{\frac{1}{1-2 x \cos \alpha+x^2 \cos ^2 \alpha+x^2 \sin ^2 \alpha}}{(1-x \cos \alpha)^2} \\ & \times \frac{(1-x \cos \alpha) \sin \alpha+(x \sin \alpha) \cos \alpha}{(1-x \cos \alpha)^{2^{\prime}}} \\ & =\frac{\sin \alpha-x \sin \alpha \cos \alpha+x \sin \alpha \cos \alpha}{1+2(-\cos \alpha) x+x^2} \\ & =\frac{\sin \alpha}{x^2+2(-\cos \alpha)+1} \\ & =\frac{\mathrm{m}}{x^2+2 \mathrm{n} x+1} \\ & \Rightarrow \mathrm{n}=-\cos \alpha \text { and } \mathrm{m}=\sin \alpha \\ & \Rightarrow m^2+n^2=1 \\ & \end{aligned}\)