MHT CET · Maths · Trigonometric Equations
The principal solutions of \(\cot x=\sqrt{3}\) are
- 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 \(\frac{\pi}{3}, \frac{7 \pi}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)
Step-by-step Solution
Detailed explanation
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 \(\tan (\pi+\theta)=\tan \theta\)
\(\therefore \tan \frac{\pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{7 \pi}{6}\)
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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