Prove the following trigonometric identities.

Solution:

Given that,

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

 

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

We have to prove $x^{2}-y^{2}=a^{2}-b^{2}$

 

We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$

So,

$x^{2}-y^{2}$

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

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

 

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

$=a^{2}-b^{2}$

Hence proved.