Prove the following identities: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} $

Given:

\( x=a \sec \theta+b \tan \theta \) and \( y=a \tan \theta+b \sec \theta \)

To do:

We have to prove that \( x^{2}-y^{2}=a^{2}-b^{2} \).

Solution:

We know that,

$\sec^2 A-\tan^2 A=1$

Therefore,

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

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

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

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

$=a^2-b^2$

Hence proved.