$\begin{aligned} & ext { If } x: y: z= an \left(\frac{\pi}{15}+\alpha ight): an \left(\frac{\pi}{15}+\beta ight): \ & an \left(\frac{\pi}{15}+\gamma ight), \operatorname{then} \frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \ & \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin…

Question

$\begin{aligned} & 	ext { If } x: y: z=	an \left(\frac{\pi}{15}+\alphaight): 	an \left(\frac{\pi}{15}+\betaight): \ & 	an \left(\frac{\pi}{15}+\gammaight), \operatorname{then} \frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \ & \sin ^2(\alpha-\beta)+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)=\end{aligned}$

Aicharya · Accepted Answer

Given $\begin{aligned} & x: y: z=	an \left(\frac{\pi}{15}+\alphaight): 	an \left(\frac{\pi}{15}+\betaight): 	an \left(\frac{\pi}{15}+\gammaight) \ & 	herefore x=k 	an \left(\frac{\pi}{15}+\alphaight), y=k 	an \left(\frac{\pi}{15}+\betaight) \ & z=k 	an \left(\frac{\pi}{15}+\gammaight)\end{aligned}$ $\begin{aligned} & 	ext { Now, } \frac{z+x}{z-x} \sin ^2(\gamma-\alpha) \ & =\frac{	an \left(12^{\circ}+\gammaight)+	an \left(12^{\circ}+\alphaight)}{	an \left(12^{\circ}+\gammaight)-	an \left(12^{\circ}+\alphaight)} \cdot \sin ^2(\gamma-\alpha) \ & =\frac{\sin \left\{24^{\circ}+(\gamma+\alpha)ight\}}{\sin (\gamma-\alpha)} \cdot \sin ^2(\gamma-\alpha) \ & =\left[\sin 24^{\circ} \cos (\gamma+\alpha)+\cos 24^{\circ} \sin (\gamma+\alpha)ight] 	imes \sin (\gamma-\alpha) \ & =\begin{array}{r}\sin 24^{\circ}[\cos (\gamma+\alpha) \sin (\gamma-\alpha)] \ \quad+\cos 24^{\circ}[\sin (\gamma+\alpha) \sin (\gamma-\alpha)]\end{array}\end{aligned}$ By adding Eqs. (i), (ii) and (iii), we get $ \begin{aligned} & \frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta) \ & \quad+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)=0 \end{aligned} $

$\begin{aligned} & \text { If } x: y: z=\tan \left(\frac{\pi}{15}+\alpha\right): \tan…

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