The general solution of equation `3 tan(theta - 15^o) = tan(theta+15^o)` is:

Options

A. `npi+(-1)^(n)pi/4`
B. `(npi)/(2)+(-1)^(n)pi/4`
C. `(npi)/(3)+(-1)^(n)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 given equation:
\[ 3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ) \]

### Step 2: Use the Componendo and Dividendo Rule
We can rewrite the equation in the form of a ratio:
\[ \frac{3}{1} = \frac{\tan(\theta + 15^\circ)}{\tan(\theta - 15^\circ)} \]
Using the Componendo and Dividendo rule, we can express this as:
\[ \frac{3 + 1}{3 - 1} = \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} \]
This simplifies to:
\[ \frac{4}{2} = \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} \]
Thus, we have:
\[ 2 = \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} \]

### Step 3: Express Tangents in Terms of Sine and Cosine
Using the identity \( \tan A = \frac{\sin A}{\cos A} \), we can rewrite the tangents:
\[ \tan(\theta + 15^\circ) = \frac{\sin(\theta + 15^\circ)}{\cos(\theta + 15^\circ)} \]
\[ \tan(\theta - 15^\circ) = \frac{\sin(\theta - 15^\circ)}{\cos(\theta - 15^\circ)} \]

### Step 4: Substitute and Simplify
Substituting these into our equation gives:
\[ 2 = \frac{\frac{\sin(\theta + 15^\circ)}{\cos(\theta + 15^\circ)} + \frac{\sin(\theta - 15^\circ)}{\cos(\theta - 15^\circ)}}{\frac{\sin(\theta + 15^\circ)}{\cos(\theta + 15^\circ)} - \frac{\sin(\theta - 15^\circ)}{\cos(\theta - 15^\circ)}} \]
After simplifying, we find:
\[ 2 = \frac{\sin(\theta + 15^\circ) \cos(\theta - 15^\circ) + \sin(\theta - 15^\circ) \cos(\theta + 15^\circ)}{\sin(\theta + 15^\circ) \cos(\theta - 15^\circ) - \sin(\theta - 15^\circ) \cos(\theta + 15^\circ)} \]

### Step 5: Use the Sine Addition and Subtraction Formulas
Using the sine addition and subtraction formulas:
\[ \sin A \cos B + \sin B \cos A = \sin(A + B) \]
\[ \sin A \cos B - \sin B \cos A = \sin(A - B) \]
This gives us:
\[ 2 = \frac{\sin(2\theta)}{\sin(30^\circ)} \]
Since \( \sin(30^\circ) = \frac{1}{2} \), we can write:
\[ 2 = \frac{\sin(2\theta)}{\frac{1}{2}} \implies \sin(2\theta) = 1 \]

### Step 6: Solve for \( 2\theta \)
The equation \( \sin(2\theta) = 1 \) implies:
\[ 2\theta = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 7: Solve for \( \theta \)
Dividing both sides by 2 gives:
\[ \theta = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final General Solution
Thus, the general solution for the equation \( 3 \tan(\theta - 15^\circ) = \tan(\theta + 15^\circ) \) is:
\[ \theta = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]