How do you simplify the expression $1+{\tan }^{2}x$?

Simplifying $1+{\tan }^{2}x$:

We know that $\tan x=\frac{\sin x}{\cos x}$.

Thus, on substituting $\tan x=\frac{\sin x}{\cos x}$ in $1+{\tan }^{2}x$, we get:

$$\begin{gathered} 1+{\tan }^{2}x=1+\frac{{\sin }^{2}x}{{\cos }^{2}x} \\ =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x} \\ =\frac{1}{{\cos }^{2}x}\left[\because {\cos }^{2}x+{\sin }^{2}x=1\right] \\ =se{c}^{2}x\left[\because \frac{1}{\cos x}=secx\right] \end{gathered}$$

Therefore, $1+{\tan }^{2}x=se{c}^{2}x$.