If x: y: z = tan left(frac{pi}{15}+alpharight | vedclass

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(C) Given that $x: y: z = 	an \left(12^{\circ}+\alphaight): 	an \left(12^{\circ}+\betaight): 	an \left(12^{\circ}+\gammaight)$.Let $x = k 	a

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(C) Given that $x: y: z = 	an \left(12^{\circ}+\alphaight): 	an \left(12^{\circ}+\betaight): 	an \left(12^{\circ}+\gammaight)$.Let $x = k 	an \left(12^{\circ}+\alphaight)$,$y = k 	an \left(12^{\circ}+\betaight)$,and $z = k 	an \left(12^{\circ}+\gammaight)$.Consider the term $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)$.Using the formula $\frac{	an A + 	an B}{	an A - 	an B} = \frac{\sin(A+B)}{\sin(A-B)}$,we get:$\frac{z+x}{z-x} = \frac{	an(12^{\circ}+\gamma) + 	an(12^{\circ}+\alpha)}{	an(12^{\circ}+\gamma) - 	an(12^{\circ}+\alpha)} = \frac{\sin(24^{\circ} + \gamma + \alpha)}{\sin(\gamma - \alpha)}$.Thus,$\frac{z+x}{z-x} \sin ^2(\gamma-\alpha) = \sin(24^{\circ} + \gamma + \alpha) \sin(\gamma - \alpha)$.Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we have:$= \frac{1}{2} [\cos(24^{\circ} + \gamma + \alpha - \gamma + \alpha) - \cos(24^{\circ} + \gamma + \alpha + \gamma - \alpha)]$$= \frac{1}{2} [\cos(24^{\circ} + 2\alpha) - \cos(24^{\circ} + 2\gamma)]$.Similarly,$\frac{x+y}{x-y} \sin ^2(\alpha-\beta) = \frac{1}{2} [\cos(24^{\circ} + 2\beta) - \cos(24^{\circ} + 2\alpha)]$ and $\frac{y+z}{y-z} \sin ^2(\beta-\gamma) = \frac{1}{2} [\cos(24^{\circ} + 2\gamma) - \cos(24^{\circ} + 2\beta)]$.Adding these three expressions,we get:$\frac{1}{2} [\cos(24^{\circ} + 2\alpha) - \cos(24^{\circ} + 2\gamma) + \cos(24^{\circ} + 2\beta) - \cos(24^{\circ} + 2\alpha) + \cos(24^{\circ} + 2\gamma) - \cos(24^{\circ} + 2\beta)] = 0$.

If $x: y: z = \tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right)$,then find the value of $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)$.

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