If `tan^(-1)(1+x)+tan^(-1)(1-x)=pi/2` then `x=?`

Options

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

Explanation

To solve the equation \( \tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2} \), we can use properties of inverse trigonometric functions.

### Step 1: Use the property of inverse tangent
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 \( b = 1 - x \). Therefore, we need to check if:
\[ (1+x)(1-x) = 1 \]

### Step 2: Expand the product
Now, expand the left-hand side:
\[ (1+x)(1-x) = 1 - x^2 \]

### Step 3: Set the equation
Now, we set the equation:
\[ 1 - x^2 = 1 \]

### Step 4: Solve for \( x \)
Subtract 1 from both sides:
\[ -x^2 = 0 \]
This implies:
\[ x^2 = 0 \]
Taking the square root of both sides gives:
\[ x = 0 \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{0} \]