Solution of `tan ^(-1) (1 + x) + tan ^(-1) ( 1- x) = (pi)/(2)` is:

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the equation \( \tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2} \]

### Step 2: Use the identity for inverse tangent
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \implies ab = 1 \]
for \( a = 1 + x \) and \( b = 1 - x \). Therefore, we can write:
\[ (1 + x)(1 - x) = 1 \]

### Step 3: Expand the left side
Expanding the left side gives:
\[ 1 - x^2 = 1 \]

### Step 4: Simplify the equation
Subtracting 1 from both sides, we have:
\[ -x^2 = 0 \]

### Step 5: Solve for \( x \)
This simplifies to:
\[ x^2 = 0 \]
Taking the square root of both sides, we find:
\[ x = 0 \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{0} \]