The number of values of `x` for which `sin^(-1)(x^2-(x^4)/3+(x^6)/9)+cos^(-1)(x^4-(x^8)/3+(x^(12))/9ddot)=pi/2,` where 0<|x|

Explanation

To solve the equation

\[ \sin^{-1}\left(x^2 - \frac{x^4}{3} + \frac{x^6}{9} + \ldots\right) + \cos^{-1}\left(x^4 - \frac{x^8}{3} + \frac{x^{12}}{9} + \ldots\right) = \frac{\pi}{2} \]

where \(0 < |x| < \sqrt{3}\), we can follow these steps:

### Step 1: Recognize the Identity
We know that

\[ \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \]

for any \(a\) in the range \([-1, 1]\). Therefore, we can set:

\[ a = x^2 - \frac{x^4}{3} + \frac{x^6}{9} + \ldots \]

and

\[ b = x^4 - \frac{x^8}{3} + \frac{x^{12}}{9} + \ldots \]

### Step 2: Find the Infinite Series Sums
Both \(a\) and \(b\) are geometric series.

For \(a\):

\[ a = x^2 \left(1 - \frac{x^2}{3} + \frac{x^4}{9} + \ldots\right) \]

This is a geometric series with first term \(x^2\) and common ratio \(-\frac{x^2}{3}\). The sum of an infinite geometric series is given by:

\[ S = \frac{a_1}{1 - r} \]

Thus,

\[ a = \frac{x^2}{1 + \frac{x^2}{3}} = \frac{3x^2}{3 + x^2} \]

For \(b\):

\[ b = x^4 \left(1 - \frac{x^4}{3} + \frac{x^8}{9} + \ldots\right) \]

This is also a geometric series with first term \(x^4\) and common ratio \(-\frac{x^4}{3}\):

\[ b = \frac{x^4}{1 + \frac{x^4}{3}} = \frac{3x^4}{3 + x^4} \]

### Step 3: Set Up the Equation
Now we have:

\[ \frac{3x^2}{3 + x^2} = \frac{3x^4}{3 + x^4} \]

### Step 4: Cross Multiply
Cross-multiplying gives:

\[ 3x^2(3 + x^4) = 3x^4(3 + x^2) \]

Expanding both sides:

\[ 9x^2 + 3x^6 = 9x^4 + 3x^6 \]

### Step 5: Simplify the Equation
Subtract \(3x^6\) from both sides:

\[ 9x^2 = 9x^4 \]

Dividing both sides by 9 (assuming \(x \neq 0\)):

\[ x^2 = x^4 \]

### Step 6: Rearranging the Equation
Rearranging gives:

\[ x^4 - x^2 = 0 \]

Factoring out \(x^2\):

\[ x^2(x^2 - 1) = 0 \]

### Step 7: Solve for \(x\)
This gives us:

\[ x^2 = 0 \quad \text{or} \quad x^2 = 1 \]

Thus,

\[ x = 0, \quad x = 1, \quad x = -1 \]

### Step 8: Consider the Given Constraints
The problem states \(0 < |x| < \sqrt{3}\). Therefore, the valid solutions are:

\[ x = 1 \quad \text{and} \quad x = -1 \]

### Conclusion
Thus, the number of values of \(x\) that satisfy the equation is:

\[ \boxed{2} \]