What is `tan^(4)A-sec^(4)A+tan^(2)A +sec^(2)A` equal to ?

Options

A. 0
B. 1
C. 2
D. `-1`

Explanation

To solve the expression \( \tan^4 A - \sec^4 A + \tan^2 A + \sec^2 A \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \tan^4 A - \sec^4 A + \tan^2 A + \sec^2 A \]

### Step 2: Factor the first part
Notice that \( \tan^4 A - \sec^4 A \) can be factored using the difference of squares:
\[ \tan^4 A - \sec^4 A = (\tan^2 A - \sec^2 A)(\tan^2 A + \sec^2 A) \]

### Step 3: Substitute the identity
We know from trigonometric identities that:
\[ \sec^2 A - \tan^2 A = 1 \]
This implies:
\[ \tan^2 A - \sec^2 A = -1 \]

### Step 4: Substitute back into the expression
Now we substitute this identity back into our expression:
\[ (\tan^2 A - \sec^2 A)(\tan^2 A + \sec^2 A) + \tan^2 A + \sec^2 A \]
Substituting \( \tan^2 A - \sec^2 A = -1 \):
\[ (-1)(\tan^2 A + \sec^2 A) + \tan^2 A + \sec^2 A \]

### Step 5: Simplify the expression
This simplifies to:
\[ -(\tan^2 A + \sec^2 A) + \tan^2 A + \sec^2 A \]
Which results in:
\[ 0 \]

### Final Answer
Thus, the expression \( \tan^4 A - \sec^4 A + \tan^2 A + \sec^2 A \) is equal to:
\[ \boxed{0} \]