Question

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

Answer 1

Given,

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

by squaring both side

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

$x^2=a^2{\sec }^2\theta +b^2{\tan }^2\theta +2ab\sec \theta \tan \theta$ ................1

And,

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

by squaring both side

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

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

subtract eq.2 from 1

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

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

$x^2-y^2=a^2\left(1\right)+b^2(-1)$

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