x sin $(\alpha +y)=\sin y ~ and ~ y'\quad =\quad \frac { m }{ \left( { x }^{ 2 }+ \right) 2nx+1 } ,$ then-

দেয়া আছে,

$\begin{aligned} & x \sin (\alpha+y)=\sin y \\ \Rightarrow & x \sin \alpha \cos y+x \cos \alpha \sin y=\sin y \\ \Rightarrow & \sin y-x \cos \alpha \sin y=x \sin \alpha \cos y \\ \Rightarrow & \sin y(1-x \cos \alpha)=x \sin \alpha \cos y\end{aligned}$

$\begin{array}{l}\Rightarrow \frac{\sin y}{\cos y}=\frac{x \sin \alpha}{1-x \cos \alpha} \\ \Rightarrow \tan y=\frac{x \sin \alpha}{1-x \cos \alpha} \\ \Rightarrow y=\tan ^{-1} \frac{x \sin \alpha}{1-x \cos \alpha}\end{array}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1+\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right)^{2}} \times \frac{\sin \alpha(1-x \cos \alpha)-x \sin \alpha \cos \alpha(-1)}{(1-x \cos \alpha)^{2}}$

$\Rightarrow y^{\prime}=\frac{1}{\frac{(1-x \cos \alpha)^{2}+(x \sin \alpha)^{2}}{(1-x \cos \alpha)^{2}}} \times \frac{\sin \alpha-x \sin \alpha \cos \alpha+x \sin \alpha \cos \alpha}{(1-x \cos \alpha)^{2}}$

$\Rightarrow y^{\prime}=\frac{(1-x \cos \alpha)^{2}}{(1-x \cos \alpha)^{2}+(x \sin \alpha)^{2}} \times \frac{\sin \alpha}{(1-x \cos \alpha)^{2}}$

$\begin{array}{l}\Rightarrow y^{\prime}=\frac{\sin \alpha}{x^{2} \cdot \sin ^{2} \alpha+1-2 x \cos \alpha+x^{2} \cos ^{2} \alpha} \\ \Rightarrow y^{\prime}=\frac{\sin \alpha}{x^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)-2 x \cos \alpha+1}\end{array}$

$\Rightarrow y^{\prime}=\frac{\sin \alpha}{x^{2}-2 x \cos \alpha+1} …(i)$

দেয়া আছে,

$y^{\prime}=\frac{m}{x^{2}+2 n x+1} …(ii)$

(i) ও (ii) তুলনা করে পাই,

$\begin{aligned} & m=\sin \alpha ; n=-\cos \alpha \\ \therefore & m^{2}+n^{2} \\ = & (\sin \alpha)^{2}+(-\cos \alpha)^{2} \\ = & \sin ^{2} \alpha+\cos ^{2} \alpha\end{aligned}$

$\begin{array}{l}=1 \\ \therefore m^{2}+n^{2}=1\end{array}$