Answer
$\sin\theta$ = $\frac{1}{\sqrt 2}$
$\cos\theta$ = $\frac{1}{\sqrt 2}$
$\tan\theta$ = 1
$\cot\theta$ = 1
$\sec\theta$ = $\sqrt 2$
$\csc\theta$ == $\sqrt 2$
Work Step by Step
Given, point $(\sqrt 2, \sqrt 2)$ is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = \sqrt 2, y = \sqrt 2$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt (\sqrt 2^{2} + \sqrt 2^{2})$
= $\sqrt (2 + 2)$
= $\sqrt (4)$ = 2
i.e. $ x = \sqrt 2, y = \sqrt 2,$ and $ r= 2$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{\sqrt 2}{2}$ = $\frac{1}{\sqrt 2}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{\sqrt 2}{2}$ = $\frac{1}{\sqrt 2}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{\sqrt 2}{\sqrt 2}$ = 1
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{\sqrt 2}{\sqrt 2}$ = 1
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{2}{\sqrt 2}$ = $\sqrt 2$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{2}{\sqrt 2}$ = $\sqrt 2$
