If `3 tan (theta - 15^@) = tan (theta + 15^@)`,
then the general value of `theta` is

Options

A. `npi + (-1)^(n) cdot (pi//4)`
B. `n(pi//2) + (-1)^(n) cdot (pi//4)`
C. `n(pi//2) + (-1)^(n) cdot (pi//6)`
D. none of these

Answer

To solve the equation \(3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ)\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ) \]

### Step 2: Use the tangent addition and subtraction formulas
Using the tangent subtraction and addition formulas:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
we can express \(\tan(\theta - 15^\circ)\) and \(\tan(\theta + 15^\circ)\):
Let \(x = \tan \theta\), then:
\[ \tan(\theta - 15^\circ) = \frac{x - \tan 15^\circ}{1 + x \tan 15^\circ} \]
\[ \tan(\theta + 15^\circ) = \frac{x + \tan 15^\circ}{1 - x \tan 15^\circ} \]

### Step 3: Substitute into the equation
Substituting these into the original equation gives:
\[ 3 \left(\frac{x - \tan 15^\circ}{1 + x \tan 15^\circ}\right) = \frac{x + \tan 15^\circ}{1 - x \tan 15^\circ} \]

### Step 4: Cross-multiply
Cross-multiplying to eliminate the fractions:
\[ 3(x - \tan 15^\circ)(1 - x \tan 15^\circ) = (x + \tan 15^\circ)(1 + x \tan 15^\circ) \]

### Step 5: Expand both sides
Expanding both sides:
\[ 3(x - \tan 15^\circ - x^2 \tan 15^\circ + x \tan^2 15^\circ) = x + x^2 \tan 15^\circ + \tan 15^\circ + x \tan^2 15^\circ \]
This simplifies to:
\[ 3x - 3\tan 15^\circ - 3x^2 \tan 15^\circ + 3x \tan^2 15^\circ = x + x^2 \tan 15^\circ + \tan 15^\circ + x \tan^2 15^\circ \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ (3x - x) + (-3x^2 \tan 15^\circ - x^2 \tan 15^\circ) + (3x \tan^2 15^\circ - x \tan^2 15^\circ) - (3\tan 15^\circ + \tan 15^\circ) = 0 \]
This simplifies to:
\[ 2x - 4\tan 15^\circ - 4x^2 \tan 15^\circ + 2x \tan^2 15^\circ = 0 \]

### Step 7: Factor out common terms
Factoring out common terms:
\[ 2x(1 + \tan^2 15^\circ) - 4\tan 15^\circ(1 + x^2) = 0 \]

### Step 8: Solve for \(x\)
Setting each factor to zero gives:
1. \(2x(1 + \tan^2 15^\circ) = 0\) which gives \(x = 0\) or \(\tan \theta = 0\) leading to \(\theta = n\pi\).
2. The second factor leads to a more complex equation which we can solve further.

### Step 9: General solution for \(\theta\)
From the first factor, we have:
\[ \theta = n\pi \]
From the second factor, we can derive:
\[ 2\theta = n\pi + (-1)^n \frac{\pi}{2} \]
Thus,
\[ \theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{4} \]

### Final Answer
The general value of \(\theta\) is:
\[ \theta = n\pi + (-1)^n \frac{\pi}{4} \]