If `A+B+C= pi ( A,B,C gt 0)` and the angle C is obtuse, then

Options

A. `tan A tan B gt 1`
B. `tan A tan B lt 1`
C. `tan A tan B =1`
D. none of these

Answer

To solve the problem step by step, we start with the given information and apply trigonometric identities.

### Step-by-Step Solution:

1. **Given Information**:
We know that \( A + B + C = \pi \) and \( A, B, C > 0 \) with angle \( C \) being obtuse.

2. **Express \( A + B \)**:
From the equation \( A + B + C = \pi \), we can express \( A + B \) as:
\[ A + B = \pi - C \]

3. **Apply the Tangent Function**:
We take the tangent of both sides:
\[ \tan(A + B) = \tan(\pi - C) \]

4. **Use the Tangent Addition Formula**:
The tangent addition formula states:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
And since \( \tan(\pi - C) = -\tan C \), we can write:
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ \tan A + \tan B = -\tan C (1 - \tan A \tan B) \]

6. **Consider the Signs**:
Since \( C \) is obtuse, \( \tan C < 0 \). Therefore, \( -\tan C > 0 \). This implies:
\[ \tan A + \tan B > 0 \]
Since \( A \) and \( B \) are both acute angles, \( \tan A > 0 \) and \( \tan B > 0 \).

7. **Combining Conditions**:
We can now conclude:
\[ 1 - \tan A \tan B > 0 \implies \tan A \tan B < 1 \]

8. **Final Conclusion**:
Thus, we find that:
\[ \tan A \tan B < 1 \]

### Final Answer:
The correct conclusion is that \( \tan A \tan B < 1 \).

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