## Precalculus (6th Edition) Blitzer

The original identity is $1+{{\tan }^{2}}x=\frac{1}{{{\cos }^{2}}x}$.
Let us consider one of the Pythagorean identities: $1+{{\tan }^{2}}x={{\sec }^{2}}x$ Now, using one of the reciprocal identities, which is $\sec x=\frac{1}{\cos x}$, we get: \begin{align} & 1+{{\tan }^{2}}x={{\sec }^{2}}x \\ & 1+{{\tan }^{2}}x=\frac{1}{{{\cos }^{2}}x} \\ \end{align}