Question

If $x=a\sec \theta ,y=b\tan \theta$, then the value of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=$

• A. $1$
• B. $-1$
• C. ${\tan }^2\theta$
• D. $cose{c}^2\theta$

Answer 1

Answer: (A)

$1$

Given, $x=a\sec \theta ,y=b\tan \theta$

We need to find value of $\frac{x^2}{a^2}-\frac{y^2}{b^2}$

Thus, the values of $\frac{x^2}{a^2}=\frac{a^2{\sec }^2\theta }{a^2}={\sec }^2\theta$

and $\frac{y^2}{b^2}=\frac{b^2{\tan }^2\theta }{b^2}={\tan }^2\theta$

Therefore, $\frac{x^2}{a^2}-\frac{y^2}{b^2}={\sec }^2\theta -{\tan }^2\theta =1$ ....$[$Since $1+{\tan }^2\theta ={\sec }^2\theta ]$