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\(x=asec\theta\)
বা, \(\cfrac{x}{a}=sec\theta\)

আবার, \(y=b tan\theta\)
বা, \(\cfrac{y}{b}=tan\theta\)

\(\therefore \cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}\)
\(=\left(\cfrac{x}{a}\right)^2-\left(\cfrac{y}{b}\right)^2\)
\(=sec^2\theta-tan^2\theta\)
\(=1\) [প্রমানিত]