Question:
If $x=b \sec ^{3} \theta$ and $y=a \tan ^{3} \theta$, prove that $\left(\frac{x}{b}\right)^{2 / 3}-\left(\frac{y}{a}\right)^{2 / 3}=1$
Solution:
$x=b \sec ^{3} \theta$
$\Rightarrow \sec ^{3} \theta=\frac{x}{b}$
$\Rightarrow \sec \theta=\left(\frac{x}{b}\right)^{\frac{1}{3}} \quad \ldots \ldots(1)$
Also,
$y=a \tan ^{3} \theta$
$\Rightarrow \tan ^{3} \theta=\frac{y}{a}$
$\Rightarrow \tan \theta=\left(\frac{y}{a}\right)^{\frac{1}{3}} \quad \ldots \ldots(2)$
We know
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$\Rightarrow\left[\left(\frac{x}{b}\right)^{\frac{1}{3}}\right]^{2}-\left[\left(\frac{y}{a}\right)^{\frac{1}{3}}\right]^{2}=1 \quad[$ Using $(1)$ and $(2)]$
$\Rightarrow\left(\frac{x}{b}\right)^{\frac{2}{3}}-\left(\frac{y}{a}\right)^{\frac{2}{3}}=1$