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,
$\frac{x+y}{x-y}{\sin }^2(\alpha -\beta )=\frac{\sin 24^{\circ }}{2}\left({\sin }^2\alpha -{\sin }^2\beta \right)$
$-\frac{\cos 24^{\circ }}{2}\left({\cos }^2\alpha -{\cos }^2\beta \right)\dots$(ii)
and $\frac{y+z}{y-z}{\sin }^2(\beta -\gamma )=\frac{\sin 24^{\circ }}{2}\left({\sin }^2\beta -{\sin }^2\gamma \right)$
$-\frac{\cos 24^{\circ }}{2}\left({\cos }^2\beta -{\cos }^2\gamma \right)\dots$ (iii)
By adding Eqs. (i), (ii) and (iii), we get
$\frac{z+x}{z-x}{\sin }^2(\gamma -\alpha )+\frac{x+y}{x-y}{\sin }^2(\alpha -\beta )$
$+\frac{y+z}{y-z}{\sin }^2(\beta -\gamma )=0$