Continuity and Differentiability — Class 12 Maths Solution

Here, $x = a\sec \theta ,$ ...(1)
and $y = b\tan \theta$ ...(2)
Differentiating (1) \& (2) w.r.t. $\theta$, we get

$\cfrac{{dx}}{{d\theta }} = a\sec \theta \tan \theta$ and $\cfrac{{dy}}{{d\theta }} = b{\sec ^2}\theta$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{dy/d\theta }}{{dx/d\theta }} = \cfrac{{b{{\sec }^2}\theta }}{{a\sec \theta \tan \theta }} = \cfrac{b}{a}\cfrac{{\sec \theta }}{{\tan \theta }} = \cfrac{b}{a}\cos ec\theta$