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

Options

A. `0`
B. `pi`
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

To solve the expression \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \), we can use the formula for the sum of inverse tangents. The formula states:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{if } ab < 1 \]
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) + \pi \quad \text{if } ab > 1 \]

### Step 1: Calculate \( \tan^{-1}(1) + \tan^{-1}(2) \)

Let \( a = 1 \) and \( b = 2 \):
- Calculate \( ab = 1 \cdot 2 = 2 \) which is greater than 1.

Thus, we will use the second case of the formula:

\[ \tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{1 + 2}{1 - 1 \cdot 2}\right) + \pi \]

Calculating the values:

\[ \tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{3}{1 - 2}\right) + \pi = \tan^{-1}\left(\frac{3}{-1}\right) + \pi = \tan^{-1}(-3) + \pi \]

Using the property \( \tan^{-1}(-x) = -\tan^{-1}(x) \):

\[ \tan^{-1}(1) + \tan^{-1}(2) = -\tan^{-1}(3) + \pi \]

### Step 2: Add \( \tan^{-1}(3) \)

Now, we need to add \( \tan^{-1}(3) \):

\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = (-\tan^{-1}(3) + \pi) + \tan^{-1}(3) \]

The \( -\tan^{-1}(3) \) and \( \tan^{-1}(3) \) cancel each other out:

\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \]

### Final Answer

Thus, the final result is:

\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi \]