Question

If x=a sec o and y-b tan o, then the value of ^b2x²….a²y² is

• A. ab
• B. a²-b²
• C. a²+b²
• D. a²b²

Answer 1

The Correct Option is D

We are tasked with finding the value of \( b^2 x^2 - a^2 y^2 \) given that:

\[ x = a \sec\theta \quad \text{and} \quad y = b \tan\theta. \]

Step 1: Substitute the expressions for \( x \) and \( y \) into \( b^2 x^2 - a^2 y^2 \).

First, compute \( x^2 \) and \( y^2 \):

\[ x^2 = (a \sec\theta)^2 = a^2 \sec^2\theta, \quad y^2 = (b \tan\theta)^2 = b^2 \tan^2\theta. \]

Substitute these into \( b^2 x^2 - a^2 y^2 \):

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

Simplify:

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

Step 2: Factor out \( a^2 b^2 \).

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

Step 3: Use the trigonometric identity \( \sec^2\theta - \tan^2\theta = 1 \).

From the Pythagorean identity:

\[ \sec^2\theta - \tan^2\theta = 1. \]

Substitute this into the expression:

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

Simplify:

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

Final Answer: The value of \( b^2 x^2 - a^2 y^2 \) is \( \mathbf{a^2 b^2} \), which corresponds to option \( \mathbf{(4)} \).