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

Options

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

Answer

To solve the problem \((1 + \frac{1}{2!} + \frac{1}{4!} + \ldots)^2 - (1 + \frac{1}{3!} + \frac{1}{5!} + \ldots)^2\), we can use properties of exponential functions and their series expansions.

### Step-by-Step Solution:

1. **Identify the Series**:
The first series can be recognized as the even terms of the exponential series for \(e^x\):
\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]
The series \(1 + \frac{1}{2!} + \frac{1}{4!} + \ldots\) corresponds to \(e^1 + e^{-1}\) divided by 2:
\[ \frac{e + e^{-1}}{2} = \cosh(1) \]

2. **Rewrite the First Series**:
Thus, we can express the first series as:
\[ \left(\frac{e + e^{-1}}{2}\right)^2 \]

3. **Identify the Second Series**:
The second series consists of the odd terms of the exponential series:
\[ 1 + \frac{1}{3!} + \frac{1}{5!} + \ldots \]
This corresponds to \(e^1 - e^{-1}\) divided by 2:
\[ \frac{e - e^{-1}}{2} = \sinh(1) \]

4. **Rewrite the Second Series**:
Therefore, we can express the second series as:
\[ \left(\frac{e - e^{-1}}{2}\right)^2 \]

5. **Combine the Two Series**:
Now we can substitute these back into our original expression:
\[ \left(\frac{e + e^{-1}}{2}\right)^2 - \left(\frac{e - e^{-1}}{2}\right)^2 \]

6. **Use the Difference of Squares**:
We can simplify this using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\):
\[ \frac{1}{4} \left((e + e^{-1}) - (e - e^{-1})\right)\left((e + e^{-1}) + (e - e^{-1})\right) \]

7. **Simplify the Expressions**:
The first part simplifies to:
\[ (e + e^{-1}) - (e - e^{-1}) = 2e^{-1} \]
The second part simplifies to:
\[ (e + e^{-1}) + (e - e^{-1}) = 2e \]

8. **Final Calculation**:
Thus, we have:
\[ \frac{1}{4} \cdot (2e^{-1}) \cdot (2e) = \frac{1}{4} \cdot 4 = 1 \]

### Conclusion:
The final result is:
\[ \boxed{1} \]