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

Correct answer: Option (B): $2n\pi \pm \frac{\pi}{3}$

Given that $a$, $b$, and $c$ are in GP, we know that: 

$b^2 = ac$

Also given: $\tan\theta \cdot \sin\theta \cdot \frac{1}{6} = \cos^2\theta$

We use the identity: $\tan\theta = \frac{\sin\theta}{\cos\theta}$

$\Rightarrow \frac{\sin\theta}{\cos\theta} \cdot \sin\theta \cdot \frac{1}{6} = \cos^2\theta$

$\Rightarrow \frac{\sin^2\theta}{6\cos\theta} = \cos^2\theta$

Multiplying both sides by $6\cos\theta$:

$\Rightarrow \sin^2\theta = 6\cos^3\theta$

Now using the identity $\sin^2\theta = 1 - \cos^2\theta$:

$\Rightarrow 1 - \cos^2\theta = 6\cos^3\theta$

Rewriting:

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

Trying $\cos\theta = \frac{1}{2}$:

$6 \cdot \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$ ✅

So, $\cos\theta = \frac{1}{2}$ is a solution.

General solution: $\theta = 2n\pi \pm \frac{\pi}{3}$

Answer 2

Given that $\sin \theta$, $\cos \theta$, and $\tan \theta$ form a geometric progression (G.P.), we start with:

$\cos^2 \theta = \frac{1}{6} \sin \theta \tan \theta$ 

$\Rightarrow \cos^2 \theta = \frac{1}{6} \cdot \sin^2 \theta \cdot \frac{1}{\cos \theta}$

$\Rightarrow \cos^3 \theta = \frac{\sin^2 \theta}{6}$

Now use the identity: $\sin^2 \theta = 1 - \cos^2 \theta$

$\Rightarrow 6 \cos^3 \theta = 1 - \cos^2 \theta$

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

Factor the cubic:

$\Rightarrow (\cos \theta - \frac{1}{2})(6 \cos^2 \theta + 4 \cos \theta + 2) = 0$

$\Rightarrow 2 (\cos \theta - \frac{1}{2})(3 \cos^2 \theta + 2 \cos \theta + 1) = 0$

Solving: $\cos \theta = \frac{1}{2} \Rightarrow \cos \theta = \cos \frac{\pi}{3}$

Hence, general solution: $\theta = 2n\pi \pm \frac{\pi}{3}$