`(1+tanalphatanbeta)^2+(tanalpha-tanbeta)^2=`

Options

A. `tan^2alphatan^2beta`
B. `sec^2alphasec^2beta`
C. `tan^2alphacot^2beta`
D. `sec^2alphacos^2beta`

Explanation

To solve the expression \((1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2\), we will use algebraic identities and trigonometric identities. Let's break it down step by step.

### Step 1: Expand the squares
We start by expanding both squares in the expression using the identities for \((A + B)^2\) and \((A - B)^2\).

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

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

### Step 2: Combine the expanded expressions
Now we combine the two expanded expressions:

\[ (1 + \tan \alpha \tan \beta)^2 + (\tan \alpha - \tan \beta)^2 = \left(1 + 2 \tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta\right) + \left(\tan^2 \alpha - 2 \tan \alpha \tan \beta + \tan^2 \beta\right) \]

### Step 3: Simplify the expression
Now, we will combine like terms:

\[ = 1 + 2 \tan \alpha \tan \beta - 2 \tan \alpha \tan \beta + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta \]

The \(2 \tan \alpha \tan \beta\) and \(-2 \tan \alpha \tan \beta\) cancel out:

\[ = 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta \]

### Step 4: Factor the expression
Now, we can factor out \(1 + \tan^2 \alpha\) and \(1 + \tan^2 \beta\):

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

### Step 5: Use trigonometric identities
Using the identity \(1 + \tan^2 \theta = \sec^2 \theta\):

\[ = \sec^2 \alpha + \sec^2 \beta \]

### Final Result
Thus, the final result is:

\[ \sec^2 \alpha \sec^2 \beta \]