Answer
According to the Pythagoras identities, ${{\sin }^{2}}t+{{\cos }^{2}}t=1$, $1+{{\tan }^{2}}t={{\sec }^{2}}t$ and $1+{{\cot }^{2}}t={{\csc }^{2}}t$.
Work Step by Step
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 $1+{{\cot }^{2}}t={{\csc }^{2}}t$.
