If $ an 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

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If $	an 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

Aicharya · Accepted Answer

$\begin{aligned} & 	an y=\frac{x \sin \alpha}{1-x \cos \alpha} \ & 	herefore \quad y=	an ^{-1}\left(\frac{x \sin \alpha}{1-x \cos \alpha}ight) \ & 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+\left(\frac{x \sin \alpha}{1-x \cos \alpha}ight)^2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x \sin \alpha}{1-x \cos \alpha}ight) \ & =\frac{\frac{1}{1-2 x \cos \alpha+x^2 \cos ^2 \alpha+x^2 \sin ^2 \alpha}}{(1-x \cos \alpha)^2} \ & 	imes \frac{(1-x \cos \alpha) \sin \alpha+(x \sin \alpha) \cos \alpha}{(1-x \cos \alpha)^{2^{\prime}}} \ & =\frac{\sin \alpha-x \sin \alpha \cos \alpha+x \sin \alpha \cos \alpha}{1+2(-\cos \alpha) x+x^2} \ & =\frac{\sin \alpha}{x^2+2(-\cos \alpha)+1} \ & =\frac{\mathrm{m}}{x^2+2 \mathrm{n} x+1} \ & \Rightarrow \mathrm{n}=-\cos \alpha 	ext { and } \mathrm{m}=\sin \alpha \ & \Rightarrow m^2+n^2=1 \ & \end{aligned}$

If $\tan y=\frac{x \sin \alpha}{1-x \cos \alpha}$ and $\frac{\mathrm{d} y}{\mathrm{~d}…

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