Answer
$\sin\theta$ = $ \frac{n}{\sqrt (m^{2} + n^{2})}$
$\cos\theta$ =$ \frac{m}{\sqrt (m^{2} + n^{2})}$
$\tan\theta$ =$ \frac{n}{m}$
$\cot\theta$ =$ \frac{m}{n}$
$\sec\theta$ =$ \frac{\sqrt (m^{2} + n^{2})}{m}$
$\csc\theta$ =$ \frac{\sqrt (m^{2} + n^{2})}{n}$
Work Step by Step
Given, point (m, n) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = m, y = n$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt (m^{2} + n^{2})$
i.e. $ x = m, y = n,$ and $ r= \sqrt (m^{2} + n^{2})$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{n}{\sqrt (m^{2} + n^{2})}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{m}{\sqrt (m^{2} + n^{2})}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{n}{m}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{m}{n}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{\sqrt (m^{2} + n^{2})}{m}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{\sqrt (m^{2} + n^{2})}{n}$
