If $630 ° < A < 720 °$ and $| \tan A | = \frac{12}{5}$ then $\tan \frac{A}{2} =$

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- \frac{3}{2}

- \frac{2}{3}

\frac{2}{3}

\frac{3}{2}

answer is D.

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Detailed Solution

Given

| \tan A | = \frac{12}{5}

$\Rightarrow 630 ° < A < 720 ° i . e A \in Q_{4}$

$\tan A = - \frac{12}{5}$

$\tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^{2} \frac{A}{2}}$

$\Rightarrow - \frac{12}{5} = \frac{2 x}{1 - x^{2}} [\tan \frac{A}{2} = x]$

$\Rightarrow - 12 + 12 x^{2} = 10 x$

$\Rightarrow 12 x^{2} - 10 x - 12 = 0$

$\Rightarrow 6 x^{2} - 5 x - 6 = 0$

$\Rightarrow 6 x^{2} - 9 x + 4 x - 6 = 0$

$\Rightarrow 3 x [2 x - 3] + 2 [2 x - 3] = 0$

$\Rightarrow [3 x + 2] [2 x - 3] = 0$

$\Rightarrow x = - \frac{2}{3}, x = \frac{3}{2}$

$\Rightarrow \tan \frac{A}{2} = - \frac{2}{3}$

$∵ 315 < \frac{A}{2} < 360$

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