In a triangle ABC, if A = `tan^(-1) 2 and B = tan^(-1) 3`, then C is equal to

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Explanation

To find the angle \( C \) in triangle \( ABC \) where \( A = \tan^{-1}(2) \) and \( B = \tan^{-1}(3) \), we can follow these steps:

### Step 1: Identify the angles
Given:
- \( A = \tan^{-1}(2) \)
- \( B = \tan^{-1}(3) \)

### Step 2: Find the tangent values
From the definitions of the inverse tangent:
- \( \tan A = 2 \)
- \( \tan B = 3 \)

### Step 3: Use the angle sum identity for tangent
We know that in a triangle, the sum of the angles is \( \pi \) (or \( 180^\circ \)):
\[ A + B + C = \pi \]
This can be rearranged to find \( C \):
\[ C = \pi - (A + B) \]

### Step 4: Calculate \( A + B \)
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values:
\[ \tan(A + B) = \frac{2 + 3}{1 - (2)(3)} = \frac{5}{1 - 6} = \frac{5}{-5} = -1 \]

### Step 5: Find \( A + B \)
Since \( \tan(A + B) = -1 \), we know:
\[ A + B = \tan^{-1}(-1) \]
The angle whose tangent is \(-1\) is:
\[ A + B = \frac{3\pi}{4} \quad (\text{since } \tan \frac{3\pi}{4} = -1) \]

### Step 6: Substitute back to find \( C \)
Now we substitute \( A + B \) back into the equation for \( C \):
\[ C = \pi - \frac{3\pi}{4} = \frac{\pi}{4} \]

### Conclusion
Thus, the angle \( C \) is:
\[ C = \frac{\pi}{4} \]

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