Question

If \(\frac{1}{6}\)sinθ.cosθ.tanθ are in G.P, then the solution set θ is

• A. \(2n\pi\pm\frac{\pi}{6}\)
• B. \(2n\pi\pm\frac{\pi}{3}\)
• C. \(n\pi+(-1)^n\frac{\pi}{3}\)
• D. \(n\pi+\frac{\pi}{3}\)

Answer 1

The Correct Option is B

The correct answer is option (B): \(2n\pi\pm\frac{\pi}{3}\)

According to the given condition, if a,b and c are in GP,we have b²=ac

Similarly, 

\(tan\theta sin\theta\times\frac{1}{6}=cos^2\theta\)

\(\Rightarrow sin^2\theta=6cos^3\theta\)

\(\Rightarrow 6cos^3\theta+cos^2\theta-1=0\)

Since \(cos\theta=\frac{1}{2}\)​, the general solution becomes \(2n\pi\pm\frac{\pi}{3}\)
 

Answer 2

Given that sin θ, cos θ, and tan θ form a geometric progression (G.P.), we start with:
\( \cos^2 \theta = \frac{1}{6} \sin \theta \tan \theta \)
\( \cos^2 \theta = \frac{1}{6} \sin^2 \theta \cos \theta \)
\( \cos^3 \theta = \frac{\sin^2 \theta}{6} \)
\( 6 \cos^3 \theta = 1 - \cos^2 \theta \)
\( 6 \cos^3 \theta + \cos^2 \theta - 1 = 0 \)
This simplifies to \( (\cos \theta - \frac{1}{2})(6 \cos^2 \theta + 4 \cos \theta + 2) = 0 \)
Further simplification gives \( 2 (\cos \theta - \frac{1}{2})(3 \cos^2 \theta + 2 \cos \theta + 1) = 0 \)
Solving \( \cos \theta - \frac{1}{2} = 0 \) leads to \( \cos \theta = \frac{1}{2} \).
Therefore, \( \cos \theta = \cos \frac{\pi}{3} \).
Hence, \( \theta = 2n\pi \pm \frac{\pi}{3} \).