If `cot^(-1)(x-(x^(2))/(2)+(x^(3))/(4)-…………..)+tan^(-1)(x^(2)-(x^(4))/(2)+(x^(6))/(4)-……..)=(pi)/(2)" for "0lt |x| lt sqrt2`, then x is equal to

Options

A. 0 only
B. 1 only
C. 0, 1 both
D. None of these

Answer

To solve the equation

\[ \cot^{-1}\left(x - \frac{x^2}{2} + \frac{x^3}{4} - \ldots\right) + \tan^{-1}\left(x^2 - \frac{x^4}{2} + \frac{x^6}{4} - \ldots\right) = \frac{\pi}{2} \]

for \(0 < |x| < \sqrt{2}\), we will first analyze the series involved in both the cotangent and tangent inverse functions.

### Step 1: Identify the series for \(\cot^{-1}\)

The series for \(\cot^{-1}\) can be expressed as:

\[ S_1 = x - \frac{x^2}{2} + \frac{x^3}{4} - \ldots \]

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

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

Substituting the values:

\[ S_1 = \frac{x}{1 - \left(-\frac{x}{2}\right)} = \frac{x}{1 + \frac{x}{2}} = \frac{x}{\frac{2 + x}{2}} = \frac{2x}{2 + x} \]

### Step 2: Identify the series for \(\tan^{-1}\)

The series for \(\tan^{-1}\) can be expressed as:

\[ S_2 = x^2 - \frac{x^4}{2} + \frac{x^6}{4} - \ldots \]

This series has the first term \(a = x^2\) and the common ratio \(r = -\frac{x^2}{2}\). Using the same formula for the sum of an infinite series:

\[ S_2 = \frac{x^2}{1 - \left(-\frac{x^2}{2}\right)} = \frac{x^2}{1 + \frac{x^2}{2}} = \frac{x^2}{\frac{2 + x^2}{2}} = \frac{2x^2}{2 + x^2} \]

### Step 3: Substitute the series sums into the equation

Now substituting \(S_1\) and \(S_2\) back into the original equation:

\[ \cot^{-1}\left(\frac{2x}{2 + x}\right) + \tan^{-1}\left(\frac{2x^2}{2 + x^2}\right) = \frac{\pi}{2} \]

### Step 4: Use the identity for cotangent and tangent

Using the identity \(\tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2}\), we can set:

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

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

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

Expanding both sides:

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

### Step 6: Cancel and rearrange

Cancelling \(2x^3\) from both sides:

\[ 4x = 4x^2 \]

Dividing both sides by 4:

\[ x = x^2 \]

### Step 7: Solve for \(x\)

Rearranging gives:

\[ x^2 - x = 0 \]

Factoring out \(x\):

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

Thus, \(x = 0\) or \(x = 1\). Since we are given \(0 < |x| < \sqrt{2}\), we discard \(x = 0\) and keep:

\[ x = 1 \]

### Conclusion

The value of \(x\) that satisfies the equation is:

\[ \boxed{1} \]