If $t g^{2} α = 3$ $tg^2alpha=3$ then ${sin}^{2} (2 α)$ $sin^2(2alpha)$ equal ...?

Question

If $t g^{2} α = 3$ $tg^2alpha=3$ then ${sin}^{2} (2 α)$ $sin^2(2alpha)$ equal ...?

A) $\frac{3}{4}$ $3/4$
B) $\frac{1}{2}$ $1/2$
C) $\frac{1}{4}$ $1/4$
D) $\frac{1}{8}$ $1/8$

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Answer 1 · 2017-04-10T14:42:34

Given

${tan}^{2} α = 3$ $tan^2alpha=3$

$\Rightarrow tan α = \pm \sqrt{3}$ $=>tanalpha=pmsqrt3$

Now

$sin 2 α = \frac{2 tan α}{1 + {tan}^{2} α}$ $sin2alpha=(2tanalpha)/(1+tan^2alpha)$

$\Rightarrow sin 2 α = \frac{\pm 2 \sqrt{3}}{1 + {(\pm \sqrt{3})}^{2}}$ $=>sin2alpha=(pm2sqrt3)/(1+(pmsqrt3)^2)$

$\Rightarrow sin 2 α = \frac{\pm 2 \sqrt{3}}{4} = \pm \frac{\sqrt{3}}{2}$ $=>sin2alpha=(pm2sqrt3)/4=pmsqrt3/2$

So

${sin}^{2} 2 α = {(\pm \frac{\sqrt{3}}{2})}^{2} = \frac{3}{4}$ $sin^2 2alpha=(pmsqrt3/2)^2=3/4$