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

Options

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

Explanation

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. Here's the step-by-step solution:

### Step 1: Apply the tangent addition formula
We know that:
\[ \tan^{-1}(A) + \tan^{-1}(B) = \frac{\pi}{2} \implies AB = 1 \]
In our case, let \( A = \frac{a}{x} \) and \( B = \frac{b}{x} \). Therefore, we have:
\[ \frac{a}{x} \cdot \frac{b}{x} = 1 \]

### Step 2: Simplify the equation
This simplifies to:
\[ \frac{ab}{x^2} = 1 \]

### Step 3: Rearrange to find \( x^2 \)
Multiplying both sides by \( x^2 \) gives:
\[ ab = x^2 \]

### Step 4: Solve for \( x \)
Taking the square root of both sides, we find:
\[ x = \pm \sqrt{ab} \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = \sqrt{ab} \quad \text{or} \quad x = -\sqrt{ab} \]