What is `(1 -sin^2alpha) (1+tan^2alpha)` equal to?

Options

A. `sin^2 theta`
B. `cos^2 theta`
C. `tan^2 theta`
D. 1

Answer

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

### Step 1: Use the Pythagorean Identity
We know from trigonometric identities that:
\[ 1 - \sin^2 \alpha = \cos^2 \alpha \]
So we can replace \(1 - \sin^2 \alpha\) with \(\cos^2 \alpha\).

### Step 2: Substitute the Identity into the Expression
Now we can rewrite the expression:
\[ (1 - \sin^2 \alpha)(1 + \tan^2 \alpha) = \cos^2 \alpha (1 + \tan^2 \alpha) \]

### Step 3: Use the Identity for Tangent
Recall that:
\[ 1 + \tan^2 \alpha = \sec^2 \alpha \]
So we can replace \(1 + \tan^2 \alpha\) with \(\sec^2 \alpha\).

### Step 4: Substitute Again
Now we can rewrite the expression again:
\[ \cos^2 \alpha (1 + \tan^2 \alpha) = \cos^2 \alpha \cdot \sec^2 \alpha \]

### Step 5: Simplify the Expression
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
\[ \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \]
Thus:
\[ \cos^2 \alpha \cdot \sec^2 \alpha = \cos^2 \alpha \cdot \frac{1}{\cos^2 \alpha} = 1 \]

### Final Answer
Therefore, the value of \((1 - \sin^2 \alpha)(1 + \tan^2 \alpha)\) is:
\[ \boxed{1} \]