if ` x = a sec theta + b tan theta `
`y = b sec theta + a tan theta `
then `x^(2) - y^(2) ` is equal to

Options

A. `4ab sec theta tan theta`
B. ` a^(2) - b^(2)`
C. `b^(2) - a^(2)`
D. `a^(2) + b^(2)`

Answer

To solve the problem, we need to find the expression for \( x^2 - y^2 \) given:

\[ x = a \sec \theta + b \tan \theta \]
\[ y = b \sec \theta + a \tan \theta \]

### Step 1: Square both expressions

First, we will square both \( x \) and \( y \).

\[ x^2 = (a \sec \theta + b \tan \theta)^2 \]
Using the formula \( (A + B)^2 = A^2 + 2AB + B^2 \):

\[ x^2 = a^2 \sec^2 \theta + 2ab \sec \theta \tan \theta + b^2 \tan^2 \theta \]

Similarly, for \( y \):

\[ y^2 = (b \sec \theta + a \tan \theta)^2 \]
\[ y^2 = b^2 \sec^2 \theta + 2ab \sec \theta \tan \theta + a^2 \tan^2 \theta \]

### Step 2: Write the expression for \( x^2 - y^2 \)

Now, we will find \( x^2 - y^2 \):

\[ x^2 - y^2 = \left( a^2 \sec^2 \theta + 2ab \sec \theta \tan \theta + b^2 \tan^2 \theta \right) - \left( b^2 \sec^2 \theta + 2ab \sec \theta \tan \theta + a^2 \tan^2 \theta \right) \]

### Step 3: Simplify the expression

Now we can simplify this expression:

\[ x^2 - y^2 = (a^2 \sec^2 \theta - b^2 \sec^2 \theta) + (b^2 \tan^2 \theta - a^2 \tan^2 \theta) \]

This can be rearranged as:

\[ x^2 - y^2 = (a^2 - b^2) \sec^2 \theta + (b^2 - a^2) \tan^2 \theta \]

### Step 4: Factor out common terms

Notice that \( (b^2 - a^2) = -(a^2 - b^2) \):

\[ x^2 - y^2 = (a^2 - b^2) \sec^2 \theta - (a^2 - b^2) \tan^2 \theta \]

Factoring out \( (a^2 - b^2) \):

\[ x^2 - y^2 = (a^2 - b^2)(\sec^2 \theta - \tan^2 \theta) \]

### Step 5: Use the identity

Using the trigonometric identity \( \sec^2 \theta - \tan^2 \theta = 1 \):

\[ x^2 - y^2 = (a^2 - b^2)(1) \]

Thus, we have:

\[ x^2 - y^2 = a^2 - b^2 \]

### Final Answer

\[ x^2 - y^2 = a^2 - b^2 \]

---