Question

If $x=\sqrt{a}\tan \alpha$ and $y=\sqrt{a}\sec \alpha$, then $y^2–x^2=$__

Answer 1

$x=\sqrt{a}\tan \alpha$ and $y=\sqrt{a}\sec \alpha$

$\Rightarrow x^2=a{\tan }^2\alpha$ and

$y^2=a{\sec }^2\alpha \Rightarrow y^2-x^2=a({\sec }^2\alpha -{\tan }^2\alpha )=a$