Statement I If `A gt 0 , B gt 0` and `A+B= pi/3`, then the maximum value of tan A tan B is `1/3`.
Statement II If `a_(1)+a_(2)+a_(3)+...+a_(n)=k`(constant), then the value `a_(1)a_(2)a_(3)...a_(n)` is greatest when
`a_(1)=a_(2)=a_(3)=...+a_(n)`

Options

A. Both Statement I and Statement II are individually true and R is the correct explanation of Statement I.
B. Both Statement I and Statement II are individually true but Statement II is not the correct explanaton of Statement I.
C. Statement I is true but Statement II is false.
D. Statement I is false but Statement II is true.

Answer

To solve the problem, we need to analyze both statements provided and prove that the maximum value of \( \tan A \tan B \) is \( \frac{1}{3} \) when \( A + B = \frac{\pi}{3} \) and both \( A \) and \( B \) are positive.

### Step-by-Step Solution:

1. **Given Conditions**:
- \( A > 0 \)
- \( B > 0 \)
- \( A + B = \frac{\pi}{3} \)

2. **Expressing \( \tan A \tan B \)**:
- We want to find the maximum value of \( \tan A \tan B \).
- Using the identity for the tangent of a sum, we have:
\[ \tan(A + B) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
- The formula for the tangent of a sum gives us:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
- Setting this equal to \( \sqrt{3} \):
\[ \sqrt{3} = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

3. **Let \( x = \tan A \) and \( y = \tan B \)**:
- We need to maximize \( xy \) under the constraint:
\[ \sqrt{3}(1 - xy) = x + y \]
- Rearranging gives:
\[ \sqrt{3} - \sqrt{3}xy = x + y \]
- Thus:
\[ x + y + \sqrt{3}xy = \sqrt{3} \]

4. **Using AM-GM Inequality**:
- By the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{x + y}{2} \geq \sqrt{xy} \]
- This implies:
\[ x + y \geq 2\sqrt{xy} \]
- Substituting \( x + y = \sqrt{3} - \sqrt{3}xy \):
\[ \sqrt{3} - \sqrt{3}xy \geq 2\sqrt{xy} \]

5. **Setting \( z = xy \)**:
- Rearranging gives:
\[ \sqrt{3} \geq 2\sqrt{z} + \sqrt{3}z \]
- This is a quadratic inequality in terms of \( \sqrt{z} \).

6. **Finding Maximum Value**:
- To find the maximum value of \( z \), we set \( A = B \) (due to symmetry and the nature of the AM-GM inequality):
\[ A + A = \frac{\pi}{3} \implies 2A = \frac{\pi}{3} \implies A = \frac{\pi}{6} \]
- Thus:
\[ \tan A = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
- Therefore:
\[ \tan A \tan B = \tan\left(\frac{\pi}{6}\right) \tan\left(\frac{\pi}{6}\right) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

### Conclusion:
The maximum value of \( \tan A \tan B \) when \( A + B = \frac{\pi}{3} \) and both \( A \) and \( B \) are positive is indeed \( \frac{1}{3} \).