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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