Explanation
Differentiate $$x = a{\cos ^3}\theta $$ with respect to $$\theta $$,
$$\frac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta $$
Differentiate $$y = a{\sin ^3}\theta $$ with respect to $$\theta $$,
$$\frac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \cos \theta $$
Now,
$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$$
$$ = - \frac{{3a{{\sin }^2}\theta \cos \theta }}{{3a{{\cos }^2}\theta \sin \theta }}$$
$$ = - \tan \theta $$
Then,
$$1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( { - \tan \theta } \right)^2}$$
$$ = 1 + {\tan ^2}\theta $$
$$ = {\sec ^2}\theta $$
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