Find the following integral: $\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$

We are given the integral $\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$.
Using the trigonometric identities $\sec x = \frac{1}{\cos x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$,we can rewrite the integrand as:
$\int \frac{\frac{1}{\cos ^{2} x}}{\frac{1}{\sin ^{2} x}} d x = \int \frac{\sin ^{2} x}{\cos ^{2} x} d x$
$= \int \tan ^{2} x d x$
Using the identity $\tan ^{2} x = \sec ^{2} x - 1$,we get:
$= \int (\sec ^{2} x - 1) d x$
$= \int \sec ^{2} x d x - \int 1 d x$
$= \tan x - x + C$,where $C$ is an arbitrary constant.