Question

Prove the following identities:
If $x=a\sec \theta +b\tan \theta andy=a\tan \theta +b\sec \theta ,$ prove that $x^2-y^2=a^2-b^2$

Answer 1

Given that,

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

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

$LHS$

$x^2-y^2$

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

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

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

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

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

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

$=1\times (a^2-b^2)$

$=(a^2-b^2)$

Hence proved.