Question:
If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^{2} x^{2}-a^{2} y^{2}=$
(a) $a b$
(b) $a^{2}-b^{2}$
(c) $a^{2}+b^{2}$
(d) $a^{2} b^{2}$
Solution:
Given:
$x=a \sec \theta, y=b \tan \theta$
So,
$b^{2} x^{2}-a^{2} y^{2}$
$=b^{2}(a \sec \theta)^{2}-a^{2}(b \tan \theta)^{2}$
$=b^{2} a^{2} \sec ^{2} \theta-a^{2} b^{2} \tan ^{2} \theta$
$=b^{2} a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$
We know that,
$\sec ^{2} \theta-\tan ^{2} \theta=1$
Therefore,
$b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}$
Hence, the correct option is (d).
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