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The principal solutions of $\cot x=\sqrt{3}$ are
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Verified Answer
The correct answer is:
$\frac{\pi}{6}, \frac{7 \pi}{6}$
(B)
The given equation is $\cot \theta=\sqrt{3}$ which is same $\tan \theta=\frac{1}{\sqrt{3}}$.
We know that, $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$ and $\tan (\pi+\theta)=\tan \theta$
$\therefore \tan \frac{\pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{7 \pi}{6}$
The given equation is $\cot \theta=\sqrt{3}$ which is same $\tan \theta=\frac{1}{\sqrt{3}}$.
We know that, $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$ and $\tan (\pi+\theta)=\tan \theta$
$\therefore \tan \frac{\pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{7 \pi}{6}$
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