If `x = a sec theta cos phi, y = b sec theta sin phi, z = c tan theta`, then, the value of `x^2 /a^2 + y^2/b^2 -z^2/c^2` is :

Options

A. 1
B. 4
C. 9
D. 0

Answer

To solve the problem, we need to evaluate the expression \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} \) given the definitions of \( x \), \( y \), and \( z \):

1. **Given Equations**:
\[ x = a \sec \theta \cos \phi \]
\[ y = b \sec \theta \sin \phi \]
\[ z = c \tan \theta \]

2. **Calculate \( \frac{x^2}{a^2} \)**:
\[ \frac{x^2}{a^2} = \frac{(a \sec \theta \cos \phi)^2}{a^2} = \sec^2 \theta \cos^2 \phi \]

3. **Calculate \( \frac{y^2}{b^2} \)**:
\[ \frac{y^2}{b^2} = \frac{(b \sec \theta \sin \phi)^2}{b^2} = \sec^2 \theta \sin^2 \phi \]

4. **Calculate \( \frac{z^2}{c^2} \)**:
\[ \frac{z^2}{c^2} = \frac{(c \tan \theta)^2}{c^2} = \tan^2 \theta \]

5. **Combine the results**:
Now we substitute these results into the original expression:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = \sec^2 \theta \cos^2 \phi + \sec^2 \theta \sin^2 \phi - \tan^2 \theta \]

6. **Factor out \( \sec^2 \theta \)**:
\[ = \sec^2 \theta (\cos^2 \phi + \sin^2 \phi) - \tan^2 \theta \]

7. **Use the Pythagorean identity**:
We know that \( \cos^2 \phi + \sin^2 \phi = 1 \), so:
\[ = \sec^2 \theta \cdot 1 - \tan^2 \theta \]

8. **Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \)**:
\[ = 1 \]

Thus, the final value of \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} \) is \( 1 \).

### Final Answer:
\[ \boxed{1} \]