MHT CET · Maths · Trigonometric Ratios & Identities
If angle \(\theta\) in \([0,2 \pi]\) satisfies both the equations \(\cot \theta=\sqrt{3}\) and \(\sqrt{3} \sec \theta+2=0\), then \(\theta\) is equal to
- A \(\frac{\pi}{6}\)
- B \(\frac{7 \pi}{6}\)
- C \(\frac{5 \pi}{6}\)
- D \(\frac{11 \pi}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{7 \pi}{6}\)
Step-by-step Solution
Detailed explanation
\(\cot \theta=\sqrt{3}\) and \(\sec \theta=\frac{-2}{\sqrt{3}}\) i.e., \(\tan \theta=\frac{1}{\sqrt{3}}\) and \(\cos \theta=-\frac{\sqrt{3}}{2}\)
\(\therefore \quad \theta\) lies in \(3^{\text {rd }}\) quadrant
\(\therefore \quad \theta=\frac{7 \pi}{6}\)
\(\therefore \quad \theta\) lies in \(3^{\text {rd }}\) quadrant
\(\therefore \quad \theta=\frac{7 \pi}{6}\)
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