Given the function `f(x)=(a^x+a^(-x))/2(w h e r ea >2)dotT h e nf(x+y)+f(x-y)=` `2f(x)dotf(y)` (b) `f(x)dotf(y)` `(f(x))/(f(y))` (d) none of these

Explanation

To solve the problem, we need to evaluate the expression \( f(x+y) + f(x-y) \) given the function \( f(x) = \frac{a^x + a^{-x}}{2} \) where \( a > 2 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{a^x + a^{-x}}{2} \]

2. **Calculate \( f(x+y) \)**:
\[ f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2} \]

3. **Calculate \( f(x-y) \)**:
\[ f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2} \]

4. **Add \( f(x+y) \) and \( f(x-y) \)**:
\[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)}}{2} + \frac{a^{x-y} + a^{-(x-y)}}{2} \]
Combine the fractions:
\[ = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2} \]

5. **Factor out common terms**:
Notice that \( a^{x+y} \) and \( a^{x-y} \) can be rewritten:
\[ = \frac{1}{2} \left( a^x a^y + a^{-x} a^{-y} + a^x a^{-y} + a^{-x} a^y \right) \]
Rearranging gives:
\[ = \frac{1}{2} \left( a^x (a^y + a^{-y}) + a^{-x} (a^y + a^{-y}) \right) \]

6. **Recognize the expression**:
We can factor out \( a^y + a^{-y} \):
\[ = \frac{1}{2} (a^y + a^{-y})(a^x + a^{-x}) \]
Since \( f(x) = \frac{a^x + a^{-x}}{2} \) and \( f(y) = \frac{a^y + a^{-y}}{2} \), we can rewrite:
\[ = 2 f(x) f(y) \]

7. **Final result**:
Therefore, we have:
\[ f(x+y) + f(x-y) = 2 f(x) f(y) \]

### Conclusion:
The correct answer is \( 2f(x) \cdot f(y) \).