show that `tan^(-1)(1/2)+tan^(-1)(2/11)=tan^(-1)(3/4)`

Explanation

To show that \( \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{3}{4}\right) \), we can use the formula for the sum of inverse tangents:

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

provided that \( xy < 1 \).

### Step-by-Step Solution:

1. **Identify \( x \) and \( y \)**:
Let \( x = \frac{1}{2} \) and \( y = \frac{2}{11} \).

2. **Calculate \( x + y \)**:
\[ x + y = \frac{1}{2} + \frac{2}{11} \]
To add these fractions, find a common denominator:
\[ = \frac{11}{22} + \frac{4}{22} = \frac{15}{22} \]

3. **Calculate \( xy \)**:
\[ xy = \left(\frac{1}{2}\right) \left(\frac{2}{11}\right) = \frac{2}{22} = \frac{1}{11} \]

4. **Check if \( xy < 1 \)**:
Since \( \frac{1}{11} < 1 \), we can use the formula.

5. **Calculate \( 1 - xy \)**:
\[ 1 - xy = 1 - \frac{1}{11} = \frac{11}{11} - \frac{1}{11} = \frac{10}{11} \]

6. **Apply the formula**:
Using the sum of inverse tangents formula:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{\frac{15}{22}}{\frac{10}{11}}\right) \]

7. **Simplify the right-hand side**:
\[ \frac{\frac{15}{22}}{\frac{10}{11}} = \frac{15}{22} \times \frac{11}{10} = \frac{15 \times 11}{22 \times 10} = \frac{165}{220} = \frac{3}{4} \]

8. **Conclusion**:
Therefore, we have shown that:
\[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]