1 + tan²x = sec²x

Divide every term of ${\sin\left(x\right)}^{2}+{\cos\left(x\right)}^{2} = 1$ by ${\cos\left(x\right)}^{2}$. Then $\frac{{\sin\left(x\right)}^{2}}{{\cos\left(x\right)}^{2}}$ is ${\tan\left(x\right)}^{2}$, $\frac{{\cos\left(x\right)}^{2}}{{\cos\left(x\right)}^{2}}$ is 1, and $\frac{1}{{\cos\left(x\right)}^{2}}$ is ${\sec\left(x\right)}^{2}$. This gives a second identity.

$$1+{\tan\left(x\right)}^{2} = {\sec\left(x\right)}^{2}$$

It is true wherever cos x is not 0. Subtract 1 from both sides: ${\sec\left(x\right)}^{2}-1 = {\tan\left(x\right)}^{2}$. So $\frac{{\sec\left(x\right)}^{2}-1}{\tan\left(x\right)}$ is $\frac{{\tan\left(x\right)}^{2}}{\tan\left(x\right)}$, which is tan x.