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

• A

  $\sin ^2 \theta$
• B

  $\cos ^2 \theta$
• Correct answer
  C

  0
• D

  1

Solution

Given $\begin{aligned} & x: y: z=\tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right) \\ & \therefore x=k \tan \left(\frac{\pi}{15}+\alpha\right), y=k \tan \left(\frac{\pi}{15}+\beta\right) \\ & z=k \tan \left(\frac{\pi}{15}+\gamma\right)\end{aligned}$ $\begin{aligned} & \text { Now, } \frac{z+x}{z-x} \sin ^2(\gamma-\alpha) \\ & =\frac{\tan \left(12^{\circ}+\gamma\right)+\tan \left(12^{\circ}+\alpha\right)}{\tan \left(12^{\circ}+\gamma\right)-\tan \left(12^{\circ}+\alpha\right)} \cdot \sin ^2(\gamma-\alpha) \\ & =\frac{\sin \left\{24^{\circ}+(\gamma+\alpha)\right\}}{\sin (\gamma-\alpha)} \cdot \sin ^2(\gamma-\alpha) \\ & =\left[\sin 24^{\circ} \cos (\gamma+\alpha)+\cos 24^{\circ} \sin (\gamma+\alpha)\right] \times \sin (\gamma-\alpha) \\ & =\begin{array}{r}\sin 24^{\circ}[\cos (\gamma+\alpha) \sin (\gamma-\alpha)] \\ \quad+\cos 24^{\circ}[\sin (\gamma+\alpha) \sin (\gamma-\alpha)]\end{array}\end{aligned}$

By adding Eqs. (i), (ii) and (iii), we get $$ \begin{aligned} & \frac{z+x}{z-x} \sin ^2(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^2(\alpha-\beta) \\ & \quad+\frac{y+z}{y-z} \sin ^2(\beta-\gamma)=0 \end{aligned} $$

Hence, the correct answer is option: 'C'.