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

### Step 1: Write down 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 inverse tangents
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \quad \text{if and only if} \quad ab = 1 \]
In our case, let \( a = 1 + x \) and \( b = 1 - x \). Therefore, we need to check if:
\[ (1 + x)(1 - x) = 1 \]

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

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

### Step 5: Solve for \( x \)
This implies:
\[ x^2 = 0 \]
Taking the square root gives:
\[ x = 0 \]

### Step 6: Determine the number of solutions
Since we found \( x = 0 \) as the only solution, we conclude that there is exactly one solution to the equation.

### Final Answer
The number of solutions of the equation is:
\[ \text{1} \]

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