What is `(1+tan alpha tan beta)^2+(tan alpha-tan beta)^2-sec^2 alpha" "sec^2beta` equal to?

Options

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

Answer

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

### Step 1: Expand the squares
We start by expanding both 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 combine these expanded terms:
\[ (1 + 2 \tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta) + (\tan^2 \alpha - 2 \tan \alpha \tan \beta + \tan^2 \beta) - \sec^2 \alpha \sec^2 \beta \]

### Step 3: Simplify the expression
Combining like terms:
\[ 1 + (\tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta) + (2 \tan \alpha \tan \beta - 2 \tan \alpha \tan \beta) - \sec^2 \alpha \sec^2 \beta \]
The \( 2 \tan \alpha \tan \beta \) and \( -2 \tan \alpha \tan \beta \) cancel each other out:
\[ 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta - \sec^2 \alpha \sec^2 \beta \]

### Step 4: Use the identity for secant
Recall the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Thus, we can express \( \sec^2 \alpha \) and \( \sec^2 \beta \):
\[ \sec^2 \alpha = 1 + \tan^2 \alpha \quad \text{and} \quad \sec^2 \beta = 1 + \tan^2 \beta \]
So,
\[ \sec^2 \alpha \sec^2 \beta = (1 + \tan^2 \alpha)(1 + \tan^2 \beta) = 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta \]

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

### Step 6: Simplify further
All terms cancel out:
\[ 1 - 1 + \tan^2 \alpha - \tan^2 \alpha + \tan^2 \beta - \tan^2 \beta + \tan^2 \alpha \tan^2 \beta - \tan^2 \alpha \tan^2 \beta = 0 \]

### Final Result
Thus, the expression simplifies to:
\[ \boxed{0} \]