Answer
$\sin\theta=\dfrac{\sqrt{6}}{3}$
$\cos\theta=\dfrac{\sqrt 3}{3}$
$\tan\theta=\sqrt 2$
$\cot\theta=\dfrac{\sqrt 2}{2}$
$\sec\theta=\sqrt 3$
$\csc\theta=\dfrac{\sqrt 6}{2}$
Work Step by Step
Let's note:
$h$=the hypotenuse
$o$=the opposite side of angle $\theta$
$a$=the adjacent side of angle $\theta$
We are given:
$o=\sqrt 2$
$a=1$
Determine the hypotenuse, using the Pythagorean Theorem:
$h^2=o^2+a^2$
$h^2=(\sqrt 2)^2+1^2$
$h^2=3$
$h=\pm\sqrt{3}$
Since $h$ is never negative, then $h=\sqrt3$.
Determine the 6 trigonometric functions of angle $\theta$:
$\sin\theta=\dfrac{o}{h}=\dfrac{\sqrt 2}{\sqrt 3}=\dfrac{\sqrt{6}}{3}$
$\cos\theta=\dfrac{a}{h}=\dfrac{1}{\sqrt 3}=\dfrac{\sqrt 3}{3}$
$\tan\theta=\dfrac{o}{a}=\dfrac{\sqrt 2}{1}=\sqrt 2$
$\cot\theta=\dfrac{a}{o}=\dfrac{1}{\sqrt 2}=\dfrac{\sqrt 2}{2}$
$\sec\theta=\dfrac{h}{a}=\dfrac{\sqrt 3}{1}=\sqrt 3$
$\csc\theta=\dfrac{h}{o}=\dfrac{\sqrt 3}{\sqrt 2}=\dfrac{\sqrt 6}{2}$
