Question:
If $x=a \cos ^{3} \theta$ and $y=b \sin ^{3} \theta$, prove that $\left(\frac{x}{a}\right)^{2 / 3}+\left(\frac{y}{b}\right)^{2 / 3}=1$
Solution:
We have $x=a \cos ^{3} \theta$
$=>\frac{x}{a}=\cos ^{3} \theta \quad \ldots$ (i)
Again, $y=b \sin ^{3} \theta$
$=>\frac{y}{b}=\sin ^{3} \theta \quad \ldots$ (ii)
Now, LHS $=\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}$
$=\left(\cos ^{3} \theta\right)^{\frac{2}{3}}+\left(\sin ^{3} \theta\right)^{\frac{2}{3}} \quad[$ From (i) and (ii) $]$
$=\cos ^{2} \theta+\sin ^{2} \theta$
$=1$
Hence, LHS $=$ RHS