If `x=asinthetaandy=btantheta`, then the value of `(a^(2))/(x^(2))-(b^(2))/(y^(2))=?`

Options

A. 5
B. `-1`
C. 1
D. 3

Answer

To solve the problem, we start with the given expressions for \( x \) and \( y \):

1. **Given:**
\[ x = a \sin \theta \quad \text{and} \quad y = b \tan \theta \]

2. **Find \( x^2 \) and \( y^2 \):**
- Squaring \( x \):
\[ x^2 = (a \sin \theta)^2 = a^2 \sin^2 \theta \]
- Squaring \( y \):
\[ y^2 = (b \tan \theta)^2 = b^2 \tan^2 \theta \]

3. **Substituting \( x^2 \) and \( y^2 \) into the expression:**
We need to evaluate:
\[ \frac{a^2}{x^2} - \frac{b^2}{y^2} \]
Substituting the values we found:
\[ \frac{a^2}{x^2} = \frac{a^2}{a^2 \sin^2 \theta} = \frac{1}{\sin^2 \theta} \]
\[ \frac{b^2}{y^2} = \frac{b^2}{b^2 \tan^2 \theta} = \frac{1}{\tan^2 \theta} \]

4. **Now, substituting these into the expression:**
\[ \frac{a^2}{x^2} - \frac{b^2}{y^2} = \frac{1}{\sin^2 \theta} - \frac{1}{\tan^2 \theta} \]

5. **Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):**
We know:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Therefore:
\[ \frac{1}{\tan^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \]

6. **Now substituting this back into the expression:**
\[ \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1 - \cos^2 \theta}{\sin^2 \theta} \]

7. **Using the Pythagorean identity \( 1 - \cos^2 \theta = \sin^2 \theta \):**
\[ \frac{1 - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1 \]

8. **Final Result:**
Thus, the value of the expression is:
\[ \frac{a^2}{x^2} - \frac{b^2}{y^2} = 1 \]