`tan^(-1)((1)/(2))+tan^(-1)((2)/(11))=`

Options

A. `tan^(-1)((4)/(3))`
B. `tan^(-1)((3)/(4))`
C. `pi+tan^(-1)((4)/(3))`
D. `pi+tan^(-1)((3)/(4))`

Answer

To solve the problem \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) \), we will use the formula for the sum of inverse tangents:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \quad \text{if } xy < 1 \]

### Step 1: Identify \( x \) and \( y \)
Let:
- \( x = \frac{1}{2} \)
- \( y = \frac{2}{11} \)

### Step 2: Calculate \( xy \)
Now, we need to check if \( xy < 1 \):
\[ xy = \left(\frac{1}{2}\right) \left(\frac{2}{11}\right) = \frac{2}{22} = \frac{1}{11} \]
Since \( \frac{1}{11} < 1 \), we can use the formula.

### Step 3: Apply the formula
Now, we can apply the formula:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{2}{11}}{1 - \frac{1}{11}}\right) \]

### Step 4: Simplify the numerator
Calculate the numerator:
\[ \frac{1}{2} + \frac{2}{11} = \frac{11}{22} + \frac{4}{22} = \frac{15}{22} \]

### Step 5: Simplify the denominator
Calculate the denominator:
\[ 1 - \frac{1}{11} = \frac{11}{11} - \frac{1}{11} = \frac{10}{11} \]

### Step 6: Combine the results
Now, we substitute back into the formula:
\[ \tan^{-1}\left(\frac{\frac{15}{22}}{\frac{10}{11}}\right) = \tan^{-1}\left(\frac{15}{22} \cdot \frac{11}{10}\right) = \tan^{-1}\left(\frac{15 \cdot 11}{22 \cdot 10}\right) = \tan^{-1}\left(\frac{165}{220}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Final Result
Thus, we have:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Conclusion
The answer is \( \tan^{-1}\left(\frac{3}{4}\right) \).