What is `(1+tanalphatanbeta)^(2)+(tanalpha-tanbeta)^(2)-sec^(2)alphasec^(2)beta` equal to

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To solve the expression \((1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2 - \sec^2 \alpha \sec^2 \beta\), we will follow these steps:

### Step 1: Expand the squares
First, we will expand the squares in the expression.

1. Expand \((1 + \tan \alpha \tan \beta)^2\):
\[ (1 + \tan \alpha \tan \beta)^2 = 1 + 2\tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta \]

2. Expand \((\tan \alpha - \tan \beta)^2\):
\[ (\tan \alpha - \tan \beta)^2 = \tan^2 \alpha - 2\tan \alpha \tan \beta + \tan^2 \beta \]

### Step 2: Combine the expanded terms
Now, we will combine the expanded terms:
\[ 1 + (1 + 2\tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta) + (\tan^2 \alpha - 2\tan \alpha \tan \beta + \tan^2 \beta) - \sec^2 \alpha \sec^2 \beta \]

### Step 3: Simplify the expression
Combine like terms:
\[ 1 + 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta + 2\tan \alpha \tan \beta - 2\tan \alpha \tan \beta - \sec^2 \alpha \sec^2 \beta \]
This simplifies to:
\[ 2 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta - \sec^2 \alpha \sec^2 \beta \]

### Step 4: Substitute \(\sec^2\) in terms of \(\tan^2\)
Recall that \(\sec^2 \alpha = 1 + \tan^2 \alpha\) and \(\sec^2 \beta = 1 + \tan^2 \beta\). Thus,
\[ \sec^2 \alpha \sec^2 \beta = (1 + \tan^2 \alpha)(1 + \tan^2 \beta) = 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta \]

### Step 5: Substitute back into the expression
Now substitute this back into our simplified expression:
\[ 2 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta - (1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta) \]

### Step 6: Final simplification
This results in:
\[ 2 - 1 = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]