If `1/6 sin theta, cos theta, tan theta` are in G.P. then `theta`=

Options

A. `2 n pi pm pi/3`
B. `2n pi pm pi/6`
C. `n pi + (-1)^n pi/3`
D. `n pi + pi/3`

Answer

To solve the problem where \( \frac{1}{6} \sin \theta, \cos \theta, \tan \theta \) are in geometric progression (G.P.), we follow these steps:

### Step 1: Understand the Condition for G.P.
For three terms \( a, b, c \) to be in G.P., the condition is that \( b^2 = ac \). Here, we have:
- \( a = \frac{1}{6} \sin \theta \)
- \( b = \cos \theta \)
- \( c = \tan \theta \)

Thus, we need to set up the equation:
\[ \cos^2 \theta = \left(\frac{1}{6} \sin \theta\right) \tan \theta \]

### Step 2: Substitute \( \tan \theta \)
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting this into the equation gives:
\[ \cos^2 \theta = \left(\frac{1}{6} \sin \theta\right) \left(\frac{\sin \theta}{\cos \theta}\right) \]
This simplifies to:
\[ \cos^2 \theta = \frac{1}{6} \frac{\sin^2 \theta}{\cos \theta} \]

### Step 3: Cross Multiply
Cross multiplying gives:
\[ \cos^3 \theta = \frac{1}{6} \sin^2 \theta \]

### Step 4: Use the Pythagorean Identity
We know that \( \sin^2 \theta = 1 - \cos^2 \theta \). Substituting this into the equation:
\[ \cos^3 \theta = \frac{1}{6} (1 - \cos^2 \theta) \]

### Step 5: Rearranging the Equation
Multiplying through by 6 to eliminate the fraction:
\[ 6 \cos^3 \theta = 1 - \cos^2 \theta \]
Rearranging gives:
\[ 6 \cos^3 \theta + \cos^2 \theta - 1 = 0 \]

### Step 6: Factor the Polynomial
Let \( x = \cos \theta \). The equation becomes:
\[ 6x^3 + x^2 - 1 = 0 \]
We can try \( x = \frac{1}{2} \) as a potential root:
\[ 6\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 1 = 6 \cdot \frac{1}{8} + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0 \]
Thus, \( x = \frac{1}{2} \) is indeed a root.

### Step 7: Factor Out \( (x - \frac{1}{2}) \)
Using synthetic division or polynomial long division, we can factor \( 6x^3 + x^2 - 1 \) as:
\[ (x - \frac{1}{2})(6x^2 + 4x + 2) = 0 \]

### Step 8: Solve for \( x \)
The first factor gives:
\[ \cos \theta = \frac{1}{2} \]
This implies:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} \]

The second factor \( 6x^2 + 4x + 2 = 0 \) has no real solutions since the discriminant \( 4^2 - 4 \cdot 6 \cdot 2 < 0 \).

### Final Answer
Thus, the solutions for \( \theta \) are:
\[ \theta = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \theta = -\frac{\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} \]