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

Options

A. x=1
B. x=-1
C. x=0
D. `x=pi`

Answer

To solve the equation \( \tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2} \), we can use the property of inverse tangent functions. Here’s a step-by-step solution:

### Step 1: Use the property of inverse tangent
Recall the identity:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \frac{\pi}{2} \quad \text{if } ab = 1 \]
In our case, let \( a = 1+x \) and \( b = 1-x \).

### Step 2: Set up the equation
We need to check if \( (1+x)(1-x) = 1 \):
\[ (1+x)(1-x) = 1 - x^2 \]
Set this equal to 1:
\[ 1 - x^2 = 1 \]

### Step 3: Solve for \( x \)
From the equation \( 1 - x^2 = 1 \), we simplify:
\[ -x^2 = 0 \]
This implies:
\[ x^2 = 0 \]

### Step 4: Find the value of \( x \)
Taking the square root of both sides gives:
\[ x = 0 \]

### Conclusion
Thus, the solution to the equation \( \tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2} \) is:
\[ \boxed{0} \]