MCQ
The solution of $tan\,\, 2\theta\,\, tan\theta = 1$ is
- A$\frac{\pi }{3}$
- ✓$(6n \pm 1)\frac{\pi }{6}$
- C$(4n \pm 1)\frac{\pi }{6}$
- D$(2n \pm 1)\frac{\pi }{6}$
$\Rightarrow \quad 2 \tan ^{2} \theta=1-\tan ^{2} \theta \Rightarrow 3 \tan ^{2} \theta=1$
$\Rightarrow \quad \tan \theta=\pm \frac{1}{\sqrt{3}}=\tan \left(\pm \frac{\pi}{6}\right)$
$\Rightarrow \quad \theta=n \pi \pm \frac{\pi}{6} \quad(m \in Z)$
$=(6 n \pm 1) \frac{\pi}{6}$
$\mathrm{Or}$
$\tan 2 \theta=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right)$
$\Rightarrow \quad 2 \theta=n \pi+\frac{\pi}{2}-\theta$
$\Rightarrow \quad 3 \theta=n \pi+\frac{\pi}{2}$
$\Rightarrow \quad \theta=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{6}=(2 \mathrm{n}+1) \frac{\pi}{6}$
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Then $(4 \alpha-8)^{2}$ is equal to
$S_1=\{z \in C:|z|<4\}, S_2=\left\{z \in C: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]>0\right\} \text { and } $
$S_3:\{z \in C: \operatorname{Re} z>0\} .$
$1.$ Area of $S=$
$(A)$ $\frac{10 \pi}{3}$ $(B)$ $\frac{20 \pi}{3}$ $(C)$ $\frac{16 \pi}{3}$ $(D)$ $\frac{32 \pi}{3}$
$2.$ $\min _{z \in S}|1-3 i-z|=$
$(A)$ $\frac{2-\sqrt{3}}{2}$ $(B)$ $\frac{2+\sqrt{3}}{2}$ $(C)$ $\frac{3-\sqrt{3}}{2}$$(D)$ $\frac{3+\sqrt{3}}{2}$
Give the answer question $1$ and $2$