The value of `tan^(-1)1+tan^(-1)2+tan^(-1)3` is

Options

A. (a)`(pi)/(2)`
B. (b)`pi`
C. (c)`(3pi)/(4)`
D. (d)`(5pi)/(8)`

Answer

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

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

If \( xy > 1 \), we need to add \( \pi \) to the result.

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

Let \( x = 1 \) and \( y = 2 \):

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

Calculating the numerator and denominator:

- Numerator: \( 1 + 2 = 3 \)
- Denominator: \( 1 - 1 \cdot 2 = 1 - 2 = -1 \)

Thus,

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

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

Now we have:

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

Using the identity again:

Let \( x = -3 \) and \( y = 3 \):

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

Calculating the numerator and denominator:

- Numerator: \( -3 + 3 = 0 \)
- Denominator: \( 1 - (-3)(3) = 1 + 9 = 10 \)

Thus,

\[ \tan^{-1}(-3) + \tan^{-1}(3) = \tan^{-1}\left(\frac{0}{10}\right) = \tan^{-1}(0) \]

### Step 3: Final Result

We know that:

\[ \tan^{-1}(0) = 0 \]

However, we need to check the signs. Since \( \tan^{-1}(-3) \) is in the fourth quadrant and \( \tan^{-1}(3) \) is in the first quadrant, they cancel each other out. Thus, we have:

\[ \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1}(0) + \pi = \pi \]

### Conclusion

Therefore, the value of \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \) is \( \pi \).