Question

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) \frac{z+x}{z-x } \sin^{2}\left(\gamma- \alpha\right)+ \frac{x+y}{x-y} \sin^{2} \left(\alpha-\beta\right)+ \frac{y+z}{y-z} \sin^{2} \left(\beta - \gamma\right) = $

• A. $\sin^2 \theta$
• B. $\cos^2 \theta$
• C. 0
• D. 1

Answer 1

Answer: (C)

Given
$x: y: z=\tan \left(\frac{\pi}{15}+\alpha\right): \tan \left(\frac{\pi}{15}+\beta\right): \tan \left(\frac{\pi}{15}+\gamma\right)$
$\therefore x=k \tan \left(\frac{\pi}{15}+\alpha\right), y=k \tan \left(\frac{\pi}{15}+\beta\right)$,
$z=k \tan \left(\frac{\pi}{15}+\gamma\right)$
Now, $\frac{z+x}{z-x} \sin ^{2}(\gamma-\alpha)$
$=\frac{\tan \left(12^{\circ}+\gamma\right)+\tan \left(12^{\circ}+\alpha\right)}{\tan \left(12^{\circ}+\gamma\right)-\tan \left(12^{\circ}+\alpha\right)} \cdot \sin ^{2}(\gamma-\alpha)$
$=\frac{\sin \left\{24^{\circ}+(\gamma+\alpha)\right\}}{\sin (\gamma-\alpha)} \cdot \sin ^{2}(\gamma-\alpha)$
$=\left[\sin 24^{\circ} \cos (\gamma+\alpha)+\cos 24^{\circ} \sin (\gamma+\alpha)\right] \times \sin (\gamma-\alpha)$
$=\sin 24^{\circ}[\cos (\gamma+\alpha) \sin (\gamma-\alpha)]$
$+\cos 24^{\circ}[\sin (\gamma+\alpha) \sin (\gamma-\alpha)]$
$=\frac{\sin 24^{\circ}}{2}\left(\sin ^{2} \gamma-\sin ^{2} \alpha\right)-\frac{\cos 24^{\circ}}{2}\left(\cos ^{2} \gamma-\cos ^{2} \alpha\right) \dots$(i)
Similarly,
$\frac{x+y}{x-y} \sin ^{2}(\alpha-\beta)=\frac{\sin 24^{\circ}}{2}\left(\sin ^{2} \alpha-\sin ^{2} \beta\right) $
$-\frac{\cos 24^{\circ}}{2}\left(\cos ^{2} \alpha-\cos ^{2} \beta\right) \ldots$(ii)
and $ \frac{y+z}{y-z} \sin ^{2}(\beta-\gamma) =\frac{\sin 24^{\circ}}{2}\left(\sin ^{2} \beta-\sin ^{2} \gamma\right)$
$-\frac{\cos 24^{\circ}}{2}\left(\cos ^{2} \beta-\cos ^{2} \gamma\right) \ldots$ (iii)
By adding Eqs. (i), (ii) and (iii), we get
$\frac{z+x}{z-x} \sin ^{2}(\gamma-\alpha)+\frac{x+y}{x-y} \sin ^{2}(\alpha-\beta)$
$+\frac{y+z}{y-z} \sin ^{2}(\beta-\gamma)=0$