The number of solution of the equation `tan^(-1) (1 + x) + tan^(-1) (1 -x) = (pi)/(2)` is

Options

A. 2
B. 3
C. 1
D. 0

Explanation

To find the number of solutions for the equation

\[ \tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2}, \]

we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:

\[ \tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2}. \]

### Step 2: Use the Identity for the Sum of Inverses
We know that

\[ \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \quad \text{if} \quad ab = 1. \]

In our case, let \( a = 1 + x \) and \( b = 1 - x \). Thus, we need to check if:

\[ (1 + x)(1 - x) = 1. \]

### Step 3: Expand and Solve
Expanding the left-hand side gives:

\[ 1 - x^2 = 1. \]

Subtracting 1 from both sides results in:

\[ -x^2 = 0. \]

### Step 4: Solve for x
This simplifies to:

\[ x^2 = 0. \]

Taking the square root of both sides, we find:

\[ x = 0. \]

### Step 5: Conclusion
Since we found only one value for \( x \), the number of solutions to the equation is:

\[ \text{Number of solutions} = 1. \]