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Question: Answered & Verified by Expert
The principal solutions of $\cot x=\sqrt{3}$ are
MathematicsTrigonometric EquationsMHT CETMHT CET 2020 (19 Oct Shift 2)
Options:
  • A $\frac{\pi}{4}, \frac{5 \pi}{4}$
  • B $\frac{\pi}{6}, \frac{7 \pi}{6}$
  • C $\frac{\pi}{6}, \frac{5 \pi}{6}$
  • D $\frac{\pi}{3}, \frac{7 \pi}{3}$
Solution:
2577 Upvotes 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 $\tan (\pi+\theta)=\tan \theta$
$\therefore \tan \frac{\pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{7 \pi}{6}$

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