Question

If $\frac{1}{6}$sin$\theta$, cos$\theta$, tan$\theta$ are in G.P. then $\theta$ =

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

Answer 1

Answer: (A)

Option (a)

Answer 2

For numbers $a, b, c$ in G.P. we have $b^2 = ac$. Here,

$$a = \frac{1}{6}\sin\theta,\quad b = \cos\theta,\quad c = \tan\theta.$$

So,

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

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we get:

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

Multiply both sides by $\cos\theta$:

$$\cos^3\theta = \frac{1}{6}\sin^2\theta.$$

Multiply by 6:

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

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

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

Rearrange:

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

Let $x = \cos\theta$, then:

$$6x^3 + x^2 - 1 = 0.$$

Testing $x = \frac{1}{2}$:

$$6\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 1 = 6\left(\frac{1}{8}\right) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0.$$

Thus, $\cos\theta = \frac{1}{2}$, which gives:

$$\theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}.$$

Explanation (Minimal):
Numbers are in G.P. ⇒ $b^2 = ac$ leads to $6\cos^3\theta = \sin^2\theta$. Substitute $\sin^2\theta = 1-\cos^2\theta$ to form a cubic. $x=\frac{1}{2}$ is a solution, so $\cos\theta = \frac{1}{2}$ and hence $\theta = 2n\pi \pm \frac{\pi}{3}$.