If : `a * sec alpha+b * tan alpha=m` and : `atan alpha+b * sec alpha=n,` then : `a^(2)-b^(2) = `

Options

A. `m^(2)+n^(2)`
B. `m^(2)-n^(2)`
C. `m^(2)-a^(2)`
D. `m^(2)-b^(2)`

Answer

To solve the problem, we start with the two given equations:

1. \( a \cdot \sec \alpha + b \cdot \tan \alpha = m \) (Equation 1)
2. \( a \cdot \tan \alpha + b \cdot \sec \alpha = n \) (Equation 2)

We need to find \( a^2 - b^2 \).

### Step 1: Square both equations

First, we square Equation 1:
\[ (a \cdot \sec \alpha + b \cdot \tan \alpha)^2 = m^2 \]
Expanding this, we have:
\[ a^2 \sec^2 \alpha + b^2 \tan^2 \alpha + 2ab \sec \alpha \tan \alpha = m^2 \quad \text{(Equation 3)} \]

Next, we square Equation 2:
\[ (a \cdot \tan \alpha + b \cdot \sec \alpha)^2 = n^2 \]
Expanding this, we get:
\[ a^2 \tan^2 \alpha + b^2 \sec^2 \alpha + 2ab \tan \alpha \sec \alpha = n^2 \quad \text{(Equation 4)} \]

### Step 2: Subtract the two squared equations

Now, we subtract Equation 4 from Equation 3:
\[ (a^2 \sec^2 \alpha + b^2 \tan^2 \alpha + 2ab \sec \alpha \tan \alpha) - (a^2 \tan^2 \alpha + b^2 \sec^2 \alpha + 2ab \tan \alpha \sec \alpha) = m^2 - n^2 \]

This simplifies to:
\[ a^2 (\sec^2 \alpha - \tan^2 \alpha) + b^2 (\tan^2 \alpha - \sec^2 \alpha) = m^2 - n^2 \]

### Step 3: Use the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \)

We know from trigonometric identities that:
\[ \sec^2 \alpha - \tan^2 \alpha = 1 \]
Thus, we can substitute this into our equation:
\[ a^2 (1) + b^2 (\tan^2 \alpha - \sec^2 \alpha) = m^2 - n^2 \]
This gives us:
\[ a^2 - b^2 = m^2 - n^2 \]

### Final Result

Therefore, we conclude that:
\[ a^2 - b^2 = m^2 - n^2 \]