MHT CET · Maths · Sequences and Series
If \(\frac{1}{6} \sin \theta, \cos \theta, \tan \theta\) are in G.P., then the general solution of \(\theta\) is
- A \(2 n \pi \pm \frac{\pi}{3}, n \in \mathbb{Z}\)
- B \(\mathrm{n} \pi+\frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}\)
- C \(\mathrm{n} \pi+\frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}\)
- D \(2 \mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}\)
Answer & Solution
Correct Answer
(A) \(2 n \pi \pm \frac{\pi}{3}, n \in \mathbb{Z}\)
Step-by-step Solution
Detailed explanation
\(\cos^2 \theta = \left(\frac{1}{6} \sin \theta\right) (\tan \theta)\) \(\cos^2 \theta = \frac{1}{6} \sin \theta \cdot \frac{\sin \theta}{\cos \theta}\)
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