If `tan^(-1) x + tan^(-1)y + tan^(-1)z= pi then x + y + z` is equal to

Options

A. xyz
B. 0
C. 1
D. 2 xyz

Answer

To solve the problem, we need to find the value of \( x + y + z \) given that:

\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi \]

### Step-by-Step Solution:

1. **Use the Identity for Sum of Inverses**:
We start by using the identity for the sum of two inverse tangents:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
We can apply this identity to the first two terms:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \]

2. **Substituting into the Original Equation**:
Substitute this back into the original equation:
\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z = \pi \]

3. **Using the Identity Again**:
Now, we apply the identity again:
\[ \tan^{-1} \left( \frac{x + y}{1 - xy} \right) + \tan^{-1} z = \tan^{-1} \left( \frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z} \right) \]
Setting this equal to \( \pi \), we know that:
\[ \tan(\pi) = 0 \]

4. **Setting Up the Equation**:
Therefore, we have:
\[ \frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z} = 0 \]
For this fraction to be zero, the numerator must be zero:
\[ \frac{x + y}{1 - xy} + z = 0 \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ x + y + z = 0 \cdot (1 - xy) \implies x + y + z = -z \]

6. **Final Equation**:
By multiplying through by \( 1 - xy \) and simplifying, we arrive at:
\[ x + y + z - xyz = 0 \]
Thus, we can conclude:
\[ x + y + z = xyz \]

### Conclusion:
The value of \( x + y + z \) is equal to \( xyz \).