Question

Prove that $x^2-y^2=a^2-b^2$, if $x=a\sec \theta +b\tan \theta$ and

$y=a\tan \theta +b\sec \theta$.

Answer 1

$x=a\sec \theta +b\tan \theta$

$y=a\tan \theta +b\sec \theta$

$x^2-y^2=a^2{\sec }^2\theta +b^2{\tan }^2\theta +2ab\sec \theta \tan \theta -a^2{\tan }^2\theta -b^2{\sec }^2\theta -2ab\sec \theta \tan \theta$

$=a^2({\sec }^2\theta -{\tan }^2\theta )-b^2({\sec }^2\theta -{\tan }^2\theta )$

$=a^2-b^2$ ( As ${\sec }^2\theta -{\tan }^2\theta =1$)

$\therefore x^2-y^2=a^2-b^2$