If `tantheta=3tanphi`, then the maximum value of `tan^2(theta-phi)` is

Options

A. 1
B. `(1)/(3)`
C. 2
D. 4

Answer

To solve the problem, we need to find the maximum value of \( \tan^2(\theta - \phi) \) given that \( \tan \theta = 3 \tan \phi \).

### Step-by-Step Solution:

1. **Given Relation**:
We start with the relation:
\[ \tan \theta = 3 \tan \phi \]

2. **Using the Tangent Subtraction Formula**:
We can express \( \tan(\theta - \phi) \) using the formula:
\[ \tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} \]
Substituting \( \tan \theta = 3 \tan \phi \) into the formula gives:
\[ \tan(\theta - \phi) = \frac{3 \tan \phi - \tan \phi}{1 + 3 \tan \phi \cdot \tan \phi} = \frac{2 \tan \phi}{1 + 3 \tan^2 \phi} \]

3. **Finding \( \tan^2(\theta - \phi) \)**:
We need to find \( \tan^2(\theta - \phi) \):
\[ \tan^2(\theta - \phi) = \left( \frac{2 \tan \phi}{1 + 3 \tan^2 \phi} \right)^2 = \frac{4 \tan^2 \phi}{(1 + 3 \tan^2 \phi)^2} \]

4. **Letting \( x = \tan^2 \phi \)**:
Let \( x = \tan^2 \phi \). Then we can rewrite the expression as:
\[ \tan^2(\theta - \phi) = \frac{4x}{(1 + 3x)^2} \]

5. **Finding the Maximum Value**:
To find the maximum value of \( \frac{4x}{(1 + 3x)^2} \), we can differentiate it with respect to \( x \) and set the derivative to zero.

Let:
\[ f(x) = \frac{4x}{(1 + 3x)^2} \]

Using the quotient rule:
\[ f'(x) = \frac{(1 + 3x)^2 \cdot 4 - 4x \cdot 2(1 + 3x)(3)}{(1 + 3x)^4} \]

Setting the numerator to zero for maximization:
\[ 4(1 + 3x)^2 - 24x(1 + 3x) = 0 \]

Simplifying gives:
\[ 4(1 + 3x) - 24x = 0 \implies 4 - 12x = 0 \implies x = \frac{1}{3} \]

6. **Substituting Back to Find Maximum**:
Substitute \( x = \frac{1}{3} \) back into the expression for \( \tan^2(\theta - \phi) \):
\[ \tan^2(\theta - \phi) = \frac{4 \cdot \frac{1}{3}}{(1 + 3 \cdot \frac{1}{3})^2} = \frac{\frac{4}{3}}{(1 + 1)^2} = \frac{\frac{4}{3}}{4} = \frac{1}{3} \]

### Conclusion:
Thus, the maximum value of \( \tan^2(\theta - \phi) \) is:
\[ \frac{1}{3} \]