If `tan alpha + cot alpha=a`, then the value of `tan^(4)alpha + cot^(4)alpha` is equal to

Options

A. `a^(4)+4a^(2)+2`
B. `a^(4) -4a^(2)+2`
C. `a^(4) - 4a^(2)-2`
D. `-a^(2) + 2a^(2) + 4`

Answer

To solve the problem, we need to find the value of \( \tan^4 \alpha + \cot^4 \alpha \) given that \( \tan \alpha + \cot \alpha = a \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan \alpha + \cot \alpha = a \]

2. **Square both sides:**
\[ (\tan \alpha + \cot \alpha)^2 = a^2 \]
Using the identity \( (x + y)^2 = x^2 + 2xy + y^2 \):
\[ \tan^2 \alpha + 2 \tan \alpha \cot \alpha + \cot^2 \alpha = a^2 \]

3. **Recall that \( \tan \alpha \cot \alpha = 1 \):**
Thus, \( 2 \tan \alpha \cot \alpha = 2 \):
\[ \tan^2 \alpha + \cot^2 \alpha + 2 = a^2 \]
Rearranging gives:
\[ \tan^2 \alpha + \cot^2 \alpha = a^2 - 2 \]

4. **Now, square \( \tan^2 \alpha + \cot^2 \alpha \):**
\[ (\tan^2 \alpha + \cot^2 \alpha)^2 = (a^2 - 2)^2 \]
Expanding both sides:
\[ \tan^4 \alpha + 2 \tan^2 \alpha \cot^2 \alpha + \cot^4 \alpha = (a^2 - 2)^2 \]

5. **Substituting \( \tan^2 \alpha \cot^2 \alpha = 1 \):**
Thus, \( 2 \tan^2 \alpha \cot^2 \alpha = 2 \):
\[ \tan^4 \alpha + \cot^4 \alpha + 2 = (a^2 - 2)^2 \]

6. **Rearranging to find \( \tan^4 \alpha + \cot^4 \alpha \):**
\[ \tan^4 \alpha + \cot^4 \alpha = (a^2 - 2)^2 - 2 \]

7. **Expanding \( (a^2 - 2)^2 \):**
\[ (a^2 - 2)^2 = a^4 - 4a^2 + 4 \]
Therefore:
\[ \tan^4 \alpha + \cot^4 \alpha = a^4 - 4a^2 + 4 - 2 \]

8. **Final simplification:**
\[ \tan^4 \alpha + \cot^4 \alpha = a^4 - 4a^2 + 2 \]

### Conclusion:
The value of \( \tan^4 \alpha + \cot^4 \alpha \) is:
\[ \boxed{a^4 - 4a^2 + 2} \]