If $x = \sec t$ and $y = \tan t$, what is $\frac{dy}{dx}$?

Explanation

1. 01

   $\frac{dx}{dt} = \sec t\tan t$ and $\frac{dy}{dt} = \sec^{2} t$.

2. 02

   $\frac{dy}{dx} = \frac{\sec^{2} t}{\sec t\tan t} = \frac{\sec t}{\tan t} = \frac{1}{\sin t} = \operatorname{cosec} t$.