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

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(C) Given $	an \left(\frac{\pi}{4}+\frac{\alpha}{2}ight)=	an ^3\left(\frac{\pi}{4}+\frac{\beta}{2}ight)$.Let $	heta = \frac{\pi}{4}+\frac{\beta

Vedclass · Accepted Answer

$\frac{\sin \alpha}{\sin \beta}$

Vedclass · Answer

(C) Given $	an \left(\frac{\pi}{4}+\frac{\alpha}{2}ight)=	an ^3\left(\frac{\pi}{4}+\frac{\beta}{2}ight)$.Let $	heta = \frac{\pi}{4}+\frac{\beta}{2}$. Then $	an \left(\frac{\pi}{4}+\frac{\alpha}{2}ight) = 	an^3 	heta$.Using the identity $\cos \phi = \frac{1-	an^2(\phi/2)}{1+	an^2(\phi/2)}$,we have $\sin \beta = \sin(2	heta - \frac{\pi}{2}) = -\cos(2	heta) = -\frac{1-	an^2 	heta}{1+	an^2 	heta} = \frac{	an^2 	heta - 1}{	an^2 	heta + 1}$.Similarly,$\sin \alpha = \frac{	an^2(\frac{\pi}{4}+\frac{\alpha}{2}) - 1}{	an^2(\frac{\pi}{4}+\frac{\alpha}{2}) + 1} = \frac{	an^6 	heta - 1}{	an^6 	heta + 1}$.Substituting $	an^2 	heta = \frac{1+\sin \beta}{1-\sin \beta}$,we find $\sin \alpha = \frac{(\frac{1+\sin \beta}{1-\sin \beta})^3 - 1}{(\frac{1+\sin \beta}{1-\sin \beta})^3 + 1} = \frac{(1+\sin \beta)^3 - (1-\sin \beta)^3}{(1+\sin \beta)^3 + (1-\sin \beta)^3} = \frac{6\sin \beta + 2\sin^3 \beta}{2 + 6\sin^2 \beta} = \frac{\sin \beta(3+\sin^2 \beta)}{1+3\sin^2 \beta}$.Therefore,$\frac{3+\sin^2 \beta}{1+3\sin^2 \beta} = \frac{\sin \alpha}{\sin \beta}$.

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

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