If x sin (alpha+y)=sin y and y=frac{m}{x^2+2 | vedclass

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(A) Given,$x \sin (\alpha+y)=\sin y$ and $y=\frac{m}{x^2+2 n x+1}$.From $x \sin (\alpha+y)=\sin y$,we have $\frac{\sin (\alpha+y)}{\sin y} = \frac{1}{

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$1-n^2$

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(A) Given,$x \sin (\alpha+y)=\sin y$ and $y=\frac{m}{x^2+2 n x+1}$.From $x \sin (\alpha+y)=\sin y$,we have $\frac{\sin (\alpha+y)}{\sin y} = \frac{1}{x}$.Using the expansion $\sin (\alpha+y) = \sin \alpha \cos y + \cos \alpha \sin y$,we get $\sin \alpha \cot y + \cos \alpha = \frac{1}{x}$.Thus,$\cot y = \frac{1 - x \cos \alpha}{x \sin \alpha}$,which implies $	an y = \frac{x \sin \alpha}{1 - x \cos \alpha}$.Differentiating $y = 	an^{-1} \left( \frac{x \sin \alpha}{1 - x \cos \alpha} ight)$ with respect to $x$:$y' = \frac{1}{1 + \left( \frac{x \sin \alpha}{1 - x \cos \alpha} ight)^2} \cdot \frac{(1 - x \cos \alpha)(\sin \alpha) - (x \sin \alpha)(-\cos \alpha)}{(1 - x \cos \alpha)^2}$.Simplifying the numerator: $(1 - x \cos \alpha)(\sin \alpha) + x \sin \alpha \cos \alpha = \sin \alpha - x \sin \alpha \cos \alpha + x \sin \alpha \cos \alpha = \sin \alpha$.Simplifying the denominator: $(1 - x \cos \alpha)^2 + (x \sin \alpha)^2 = 1 - 2x \cos \alpha + x^2 \cos^2 \alpha + x^2 \sin^2 \alpha = 1 - 2x \cos \alpha + x^2$.Thus,$y' = \frac{\sin \alpha}{x^2 - 2x \cos \alpha + 1}$.Comparing this with the derivative of $y = \frac{m}{x^2 + 2nx + 1}$,we identify $m = \sin \alpha$ and $n = -\cos \alpha$.Therefore,$m^2 = \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - (-n)^2 = 1 - n^2$.

If $x \sin (\alpha+y)=\sin y$ and $y=\frac{m}{x^2+2 n x+1}$ then $m^2=$

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