If x = a sec θ + b tan θ and y = a tan θ + b sec θ,


If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 − y2) = (a2 − b2).


We have $x^{2}-y^{2}=\left[(a \sec \theta+b \tan \theta)^{2}-(a \tan \theta+b \sec \theta)^{2}\right]$

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

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

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

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

$=a^{2}-b^{2} \quad\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

Hence, $x^{2}-y^{2}=a^{2}-b^{2}$


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