`2tan^(-1)((1)/(3))+tan^(-1)((1)/(2))=`

Options

A. `(pi)/(2)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `tan^(-1)2`

Answer

To solve the equation \( 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) \), we will use the formula for the tangent of a sum of angles and the double angle formula for the inverse tangent function.

### Step-by-Step Solution:

1. **Use the Double Angle Formula for Inverse Tangent:**
The double angle formula for inverse tangent is given by:
\[ 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
Here, let \( x = \frac{1}{3} \). Therefore,
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) \]

2. **Calculate the Expression Inside the Inverse Tangent:**
First, calculate \( 1 - \left(\frac{1}{3}\right)^2 \):
\[ 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]
Now substitute this back into the formula:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) = \tan^{-1}\left(\frac{2}{3} \cdot \frac{9}{8}\right) = \tan^{-1}\left(\frac{18}{24}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

3. **Combine with the Other Inverse Tangent:**
Now we have:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{2}\right) \]
We will use the formula for the sum of two inverse tangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Here, \( a = \frac{3}{4} \) and \( b = \frac{1}{2} \).

4. **Calculate the Sum:**
First, calculate \( ab \):
\[ ab = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8} \]
Now substitute into the formula:
\[ \tan^{-1}\left(\frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}}\right) \]
Calculate \( \frac{3}{4} + \frac{1}{2} \):
\[ \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]
Calculate \( 1 - \frac{3}{8} \):
\[ 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} \]
Now substitute these values back into the formula:
\[ \tan^{-1}\left(\frac{\frac{5}{4}}{\frac{5}{8}}\right) = \tan^{-1}\left(\frac{5}{4} \cdot \frac{8}{5}\right) = \tan^{-1}(2) \]

5. **Final Result:**
Therefore, we have:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}(2) \]

### Conclusion:
The final answer is:
\[ \tan^{-1}(2) \]