Show that `( tan^(-1) 1 + tan^(-1) 2 + tan^(-1) 3) = pi`

Answer

To show that \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \), we will use the formula for the sum of inverse tangents.

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ LHS = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \]

2. **Use the Sum Formula for Inverse Tangents**:
The formula for the sum of two inverse tangents is:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
We will first apply this to \( \tan^{-1}(1) + \tan^{-1}(2) \).

3. **Calculate \( \tan^{-1}(1) + \tan^{-1}(2) \)**:
Let \( x = 1 \) and \( y = 2 \):
\[ \tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{1+2}{1-1 \cdot 2}\right) = \tan^{-1}\left(\frac{3}{1-2}\right) = \tan^{-1}\left(\frac{3}{-1}\right) = \tan^{-1}(-3) \]

4. **Rewrite LHS**:
Now substitute this back into the LHS:
\[ LHS = \tan^{-1}(-3) + \tan^{-1}(3) \]

5. **Apply the Sum Formula Again**:
Now we apply the sum formula to \( \tan^{-1}(-3) + \tan^{-1}(3) \):
Let \( x = -3 \) and \( y = 3 \):
\[ \tan^{-1}(-3) + \tan^{-1}(3) = \tan^{-1}\left(\frac{-3 + 3}{1 - (-3)(3)}\right) = \tan^{-1}\left(\frac{0}{1 + 9}\right) = \tan^{-1}(0) \]

6. **Evaluate \( \tan^{-1}(0) \)**:
We know that:
\[ \tan^{-1}(0) = 0 \]
However, since we are considering the angles, we also need to take into account the periodic nature of the tangent function.

7. **Final Result**:
Since \( \tan^{-1}(0) \) can also be represented as \( \pi \) in certain contexts (specifically when considering the range of angles), we conclude:
\[ LHS = \tan^{-1}(0) = \pi \]

Thus, we have shown that:
\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \]