`(1+x^2/(2!) +x^4 /(4!) + .....oo)^(2) - (x+x^3/(3!) +x^(5)/(5!) + ....oo)^(2) = `

Options

A. `-1`
B. 5
C. 2
D. 1

Answer

To solve the given problem step by step, we will analyze the two series and utilize the properties of exponential functions.

### Step 1: Identify the series
The first series is:
\[ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \]
This series represents the even powers of \(x\) in the expansion of \(e^x\).

The second series is:
\[ x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \]
This series represents the odd powers of \(x\) in the expansion of \(e^x\).

### Step 2: Relate the series to exponential functions
The series can be expressed using the exponential function:
- The first series can be recognized as:
\[ \frac{e^x + e^{-x}}{2} \]
This is because \(e^x + e^{-x}\) contains all even powers of \(x\).

- The second series can be recognized as:
\[ \frac{e^x - e^{-x}}{2} \]
This is because \(e^x - e^{-x}\) contains all odd powers of \(x\).

### Step 3: Square the series
Now we need to square both series:
\[ \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 \]

### Step 4: Use the difference of squares
Using the identity \(a^2 - b^2 = (a + b)(a - b)\), we can rewrite the expression:
Let \(a = \frac{e^x + e^{-x}}{2}\) and \(b = \frac{e^x - e^{-x}}{2}\).

Then:
\[ a + b = \frac{(e^x + e^{-x}) + (e^x - e^{-x})}{2} = \frac{2e^x}{2} = e^x \]
\[ a - b = \frac{(e^x + e^{-x}) - (e^x - e^{-x})}{2} = \frac{2e^{-x}}{2} = e^{-x} \]

### Step 5: Substitute back into the expression
Now substituting back, we have:
\[ (e^x)(e^{-x}) = 1 \]

### Conclusion
Thus, the final result of the expression is:
\[ \boxed{1} \]