If `tan theta + cot theta = x, ` then what is the value of `tan^(4) theta + cot^(4) theta ` ?

Options

A. `(x^(3) - 3)^(2) + 2`
B. `(x^(4) - 2 x) + 4 `
C. `x (x - 4) + 2`
D. `x^(2) (x^(2) - 4)+ 2`

Answer

To find the value of \( \tan^4 \theta + \cot^4 \theta \) given that \( \tan \theta + \cot \theta = x \), we can follow these steps:

### Step 1: Square the given equation
We start with the equation:
\[ \tan \theta + \cot \theta = x \]
Now, we square both sides:
\[ (\tan \theta + \cot \theta)^2 = x^2 \]

### Step 2: Expand the left side
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \), we can expand the left side:
\[ \tan^2 \theta + \cot^2 \theta + 2 \tan \theta \cot \theta = x^2 \]
Since \( \tan \theta \cot \theta = 1 \), we have:
\[ \tan^2 \theta + \cot^2 \theta + 2 = x^2 \]

### Step 3: Rearrange to find \( \tan^2 \theta + \cot^2 \theta \)
Rearranging the equation gives us:
\[ \tan^2 \theta + \cot^2 \theta = x^2 - 2 \]

### Step 4: Square \( \tan^2 \theta + \cot^2 \theta \)
Now, we square \( \tan^2 \theta + \cot^2 \theta \):
\[ (\tan^2 \theta + \cot^2 \theta)^2 = (x^2 - 2)^2 \]

### Step 5: Expand the squared term
Using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \) again:
\[ \tan^4 \theta + \cot^4 \theta + 2 \tan^2 \theta \cot^2 \theta = (x^2 - 2)^2 \]
Since \( \tan^2 \theta \cot^2 \theta = 1 \), we can substitute:
\[ \tan^4 \theta + \cot^4 \theta + 2 = (x^2 - 2)^2 \]

### Step 6: Rearrange to find \( \tan^4 \theta + \cot^4 \theta \)
Rearranging gives us:
\[ \tan^4 \theta + \cot^4 \theta = (x^2 - 2)^2 - 2 \]

### Step 7: Simplify the expression
Now we simplify the right side:
\[ \tan^4 \theta + \cot^4 \theta = (x^4 - 4x^2 + 4) - 2 = x^4 - 4x^2 + 2 \]

### Final Result
Thus, the value of \( \tan^4 \theta + \cot^4 \theta \) is:
\[ \tan^4 \theta + \cot^4 \theta = x^4 - 4x^2 + 2 \]