If tan left(frac{pi}{4}+frac{y}{2}right)=tan | vedclass

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(D) Given,$	an \left(\frac{\pi}{4}+\frac{y}{2}ight)=	an ^3\left(\frac{\pi}{4}+\frac{x}{2}ight)$ Using the formula $	an \left(\frac{\pi}{4}+	he

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$\sin y$

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(D) Given,$	an \left(\frac{\pi}{4}+\frac{y}{2}ight)=	an ^3\left(\frac{\pi}{4}+\frac{x}{2}ight)$ Using the formula $	an \left(\frac{\pi}{4}+	hetaight) = \frac{1+	an 	heta}{1-	an 	heta}$,we get: $\frac{1+	an \frac{y}{2}}{1-	an \frac{y}{2}} = \left(\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}}ight)^3$ Using $	an \frac{	heta}{2} = \frac{\sin 	heta}{1+\cos 	heta}$ or the identity $\frac{1+	an \frac{	heta}{2}}{1-	an \frac{	heta}{2}} = \frac{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} - \sin \frac{	heta}{2}}$,squaring gives: $\frac{1+\sin y}{1-\sin y} = \left(\frac{1+\sin x}{1-\sin x}ight)^3$ Applying the componendo and dividendo rule $\frac{a+b}{a-b} = \frac{c+d}{c-d}$: $\frac{(1+\sin y) + (1-\sin y)}{(1+\sin y) - (1-\sin y)} = \frac{(1+\sin x)^3 + (1-\sin x)^3}{(1+\sin x)^3 - (1-\sin x)^3}$ $\frac{2}{2 \sin y} = \frac{2(1+3 \sin ^2 x)}{2(3 \sin x + \sin ^3 x)}$ $\frac{1}{\sin y} = \frac{1+3 \sin ^2 x}{3 \sin x + \sin ^3 x}$ Therefore,$\sin y = \frac{3 \sin x + \sin ^3 x}{1+3 \sin ^2 x}$.

If $\tan \left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{x}{2}\right)$,then $\frac{3 \sin x+\sin ^3 x}{1+3 \sin ^2 x}=$

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