## Precalculus (6th Edition) Blitzer

According to the Pythagoras identities, ${{\sin }^{2}}t+{{\cos }^{2}}t=1$, $1+{{\tan }^{2}}t={{\sec }^{2}}t$ and $1+{{\cot }^{2}}t={{\csc }^{2}}t$.
If $t$ is a real number and $P=\left( x,y \right)$ is a point on the unit circle that corresponds to $t$, then, $\sin t=y$, $\cos t=x$ , $\tan t=\frac{y}{x},x\ne 0$ , $\csc t=\frac{1}{y},y\ne 0$ , $\sec t=\frac{1}{x},x\ne 0$ , $\cot t=\frac{x}{y},y\ne 0$ . The relationships among trigonometric functions follow from the equation of the unit circle. Consider the equation of a unit circle. ${{x}^{2}}+{{y}^{2}}=1$ Substitute $\sin t$ for $y$ and $\cos t$ for $x$. ${{\cos }^{2}}t+{{\sin }^{2}}t=1$ Consider the expression for $\tan t$. $\tan t=\frac{y}{x}$ Consider the expression for $\sec t$. $\sec t=\frac{1}{x}$ Consider the following expression. $1+{{\tan }^{2}}t$ Substitute $\frac{y}{x}$ for $\tan t$. \begin{align} & 1+{{\tan }^{2}}t=1+\frac{{{y}^{2}}}{{{x}^{2}}} \\ & =\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}} \\ & =\frac{1}{{{x}^{2}}} \\ & ={{\sec }^{2}}t \end{align} Consider the expression for $\cot t$. $\cot t=\frac{x}{y}$ Consider the expression for $\csc t$. $\csc t=\frac{1}{y}$ Consider the following expression. $1+{{\cot }^{2}}t$ Substitute $\frac{x}{y}$ for $\cot t$. \begin{align} & 1+{{\cot }^{2}}t=1+\frac{{{x}^{2}}}{{{y}^{2}}} \\ & =\frac{{{y}^{2}}+{{x}^{2}}}{{{y}^{2}}} \\ & =\frac{1}{{{y}^{2}}} \\ & ={{\csc }^{2}}t \end{align} Therefore, according to the Pythagoras identities, ${{\sin }^{2}}t+{{\cos }^{2}}t=1$, $1+{{\tan }^{2}}t={{\sec }^{2}}t$ and $1+{{\cot }^{2}}t={{\csc }^{2}}t$.