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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.
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