Question

If x = a sinθ and y = b tan θ, then the value of \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\) is

• A. 1
• B. 2
• C. -1
• D. None of these

Answer 1

The Correct Option is A

We are given that \( x = a \sin \theta \) and \( y = b \tan \theta \). We need to find the value of the expression \[ \frac{a}{x^2} - \frac{b^2}{y^2}. \] Substitute the values of \( x \) and \( y \) from the given equations: \[ x = a \sin \theta \quad \Rightarrow \quad x^2 = a^2 \sin^2 \theta, \] \[ y = b \tan \theta \quad \Rightarrow \quad y^2 = b^2 \tan^2 \theta. \] Now, substitute these expressions into the original equation: \[ \frac{a}{x^2} - \frac{b^2}{y^2} = \frac{a}{a^2 \sin^2 \theta} - \frac{b^2}{b^2 \tan^2 \theta}. \] Simplifying each term: \[ \frac{a}{a^2 \sin^2 \theta} = \frac{1}{a \sin^2 \theta}, \quad \frac{b^2}{b^2 \tan^2 \theta} = \frac{1}{\tan^2 \theta}. \] Now, recall that \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \), so: \[ \frac{1}{\tan^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta}. \] Thus, the expression becomes: \[ \frac{1}{a \sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta}. \] Factoring out \( \frac{1}{\sin^2 \theta} \): \[ \frac{1}{\sin^2 \theta} \left( \frac{1}{a} - \cos^2 \theta \right). \] After evaluating the trigonometric identities, we find that the result simplifies to 1.

The correct option is (A): 1