If `tan^(-1) x + tan^(-1) y + tan ^(-1) z = (pi)/(2)` , then value of xy + yz + zx a) -1 b) 1 c) 0 d) None of these

Options

A. `-1`
B. 1
C. 0
D. None of these

Answer

To solve the problem, we start with the given equation:

\[ \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2} \]

### Step 1: Assign Variables
Let:
\[ \tan^{-1} x = a, \quad \tan^{-1} y = b, \quad \tan^{-1} z = c \]
Thus, we can rewrite the equation as:
\[ a + b + c = \frac{\pi}{2} \]

### Step 2: Express x, y, z in terms of a, b, c
From the definitions of \(a\), \(b\), and \(c\), we have:
\[ x = \tan a, \quad y = \tan b, \quad z = \tan c \]

### Step 3: Use the Identity for Tangent
Using the identity for the tangent of a sum, we know:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Substituting \(c = \frac{\pi}{2} - (a + b)\), we find:
\[ \tan c = \cot(a + b) = \frac{1}{\tan(a + b)} \]

### Step 4: Substitute the Values
From our earlier expression, we can express \(\tan c\) as:
\[ \tan c = \frac{1}{\frac{\tan a + \tan b}{1 - \tan a \tan b}} = \frac{1 - \tan a \tan b}{\tan a + \tan b} \]
Substituting \(x\), \(y\), and \(z\):
\[ z = \frac{1 - xy}{x + y} \]

### Step 5: Multiply through by \(x + y\)
Now, we multiply both sides by \(x + y\):
\[ z(x + y) = 1 - xy \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ xy + yz + zx = 1 \]

### Conclusion
Thus, the value of \(xy + yz + zx\) is:
\[ \boxed{1} \]