If `tan^(-1)(a/x) + tan^(-1)(b/x) =pi/2`, then: x=

Options

A. ab
B. 2ab
C. `sqrt(ab)`
D. `sqrt(2ab)`

Answer

To solve the equation \( \tan^{-1}\left(\frac{a}{x}\right) + \tan^{-1}\left(\frac{b}{x}\right) = \frac{\pi}{2} \), we can use the property of the tangent function.

### Step-by-Step Solution:

1. **Understanding the Equation:**
We know that if \( \tan^{-1}(A) + \tan^{-1}(B) = \frac{\pi}{2} \), then \( A \cdot B = 1 \). In our case, let \( A = \frac{a}{x} \) and \( B = \frac{b}{x} \).

2. **Setting Up the Equation:**
From the property mentioned, we can write:
\[ \frac{a}{x} \cdot \frac{b}{x} = 1 \]

3. **Simplifying the Equation:**
This simplifies to:
\[ \frac{ab}{x^2} = 1 \]

4. **Solving for \( x^2 \):**
Rearranging gives:
\[ ab = x^2 \]

5. **Finding \( x \):**
Taking the square root of both sides, we find:
\[ x = \sqrt{ab} \]

Thus, the value of \( x \) is \( \sqrt{ab} \).

### Final Answer:
\[ x = \sqrt{ab} \]