## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\frac{1+\sin t}{{{\cos }^{2}}t}$, which can be simplified by using the quotient identity $\tan x=\frac{\sin x}{\cos x}$ and reciprocal identities $\frac{1}{\cos x}=\sec x$. After that, use the Pythagorean identity ${{\sec }^{2}}t=1+{{\tan }^{2}}t$ \begin{align} & \frac{1+\sin t}{{{\cos }^{2}}t}=\frac{1}{{{\cos }^{2}}t}+\frac{\sin t}{{{\cos }^{2}}t} \\ & ={{\sec }^{2}}t+\frac{\sin t}{\cos t}.\frac{1}{\cos t} \\ & ={{\sec }^{2}}t+\tan t.\sec t \\ & ={{\tan }^{2}}t+1+\tan t\sec t \end{align} Therefore, the expression on the left side is equal to the expression on the right side.