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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 22 of 160

Marks: +1, -0

If x : y : z = tan

\left(\frac{\pi}{15}+\alpha\right)

\left(\frac{\pi}{15}+\beta\right)

\left(\frac{\pi}{15}+\gamma\right)

\frac{z+x}{z-1} \sin^2

\frac{x+y}{x-y} \sin^2 (\alpha-\beta)

\frac{y+z}{y-z} \sin^2 (\beta-\gamma)

$\sin^2$ θ
$\cos^2$ θ
0
1

Solution:

x: y: z = \tan\left(\frac{\pi}{15}+\alpha\right) : \tan\left(\frac{\pi}{15}+\beta\right) : \tan\left(\frac{\pi}{15}+\gamma\right)

\text{So}

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(12^{\circ}+\gamma)+\tan(12^{\circ}+\alpha)}{\tan(12^{\circ}+\gamma)-\tan(12^{\circ}+\alpha)} \sin^{2}(\gamma-\alpha)

= \frac{\sin(24^{\circ}+(\gamma+\alpha))}{\sin(\gamma-\alpha)} \sin^{2}(\gamma-\alpha)

= [\sin 24^{\circ} \cos(\gamma-\alpha) + \cos 24^{\circ} \sin(\gamma+\alpha)] \times \sin 24^{\circ}

= \sin(\cos(\gamma+\alpha) \sin(\gamma+\alpha)) + \left[ \begin{array}{c} \cos 24^{\circ} \\ +\sin(\gamma+\alpha) \sin \alpha \end{array} \right]

Further simplify the above

\frac{\sin 24^{\circ}}{2} - (\sin^{2}\gamma - \sin^{2}\alpha) - \frac{\cos 24^{\circ}}{2} (\cos^{2}\gamma - \cos^{2}\alpha)

Similarly,

\frac{x+y}{x-y} \sin^{2}(\alpha-\beta) = \frac{\sin 24^{\circ}}{2} (\sin^{2}\alpha - \sin^{2}\beta) - \frac{\cos 24^{\circ}}{2} \left( \begin{array}{c} \cos^{2}\alpha \\ -\cos^{2}\beta \end{array} \right)

and

\frac{y+z}{y-z} \sin^{2}(\beta-\gamma) = \frac{\sin 24^{\circ}}{2} (\sin^{2}\beta - \sin^{2}\gamma) - \frac{\cos 24^{\circ}}{2} \left( \begin{array}{c} \cos^{2}\beta \\ -\cos^{2}\gamma \end{array} \right)

From above equations

\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

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