If cot A = 2 and cot B = 3, then what is the value of (A+B)?

Options

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

Answer

To solve the problem where \( \cot A = 2 \) and \( \cot B = 3 \), we need to find the value of \( A + B \). Here’s a step-by-step solution:

### Step 1: Express \( A \) and \( B \) in terms of tangent
Given that \( \cot A = 2 \), we can express \( A \) in terms of tangent:
\[ \cot A = \frac{1}{\tan A} \implies \tan A = \frac{1}{\cot A} = \frac{1}{2} \]
Thus,
\[ A = \tan^{-1}\left(\frac{1}{2}\right) \]

Similarly, for \( \cot B = 3 \):
\[ \cot B = \frac{1}{\tan B} \implies \tan B = \frac{1}{\cot B} = \frac{1}{3} \]
Thus,
\[ B = \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 2: Use the tangent addition formula
We want to find \( A + B \):
\[ A + B = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \]
We can use the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values of \( \tan A \) and \( \tan B \):
\[ \tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right)} \]

### Step 3: Simplify the numerator and denominator
Calculating the numerator:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

Calculating the denominator:
\[ 1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 4: Combine the results
Now substituting back into the tangent formula:
\[ \tan(A + B) = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \]

### Step 5: Find \( A + B \)
Since \( \tan(A + B) = 1 \), we know:
\[ A + B = \tan^{-1}(1) \]
The value of \( \tan^{-1}(1) \) is:
\[ A + B = \frac{\pi}{4} \]

### Final Answer
Thus, the value of \( A + B \) is:
\[ \frac{\pi}{4} \]

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