Answer
$\sin\theta=\dfrac{\sqrt{2}}{2}$
$\cos\theta=\dfrac{\sqrt{2}}{2}$
$\tan\theta=1$
$\cot\theta=1$
$\sec\theta=\sqrt 2$
$\csc\theta=\sqrt 2$
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:
$o=3$
$a=3$
Determine the hypotenuse, using the Pythagorean Theorem:
$h^2=o^2+a^2$
$h^2=3^2+3^2$
$h^2=18$
$h=\pm\sqrt{18}$
$h=\pm3\sqrt2$
Since $h$ is never negative, then $h=3\sqrt 2$
Determine the $6$ trigonometric functions of angle $\theta$:
$\sin\theta=\dfrac{o}{h}=\dfrac{3}{3\sqrt 2}=\dfrac{1}{\sqrt 2}=\dfrac{\sqrt{2}}{2}$
$\cos\theta=\dfrac{a}{h}=\dfrac{3}{3\sqrt 2}=\dfrac{1}{\sqrt 2}=\dfrac{\sqrt{2}}{2}$
$\tan\theta=\dfrac{o}{a}=\dfrac{3}{3}=1$
$\cot\theta=\dfrac{a}{o}=\dfrac{3}{3}=1$
$\sec\theta=\dfrac{h}{a}=\dfrac{3\sqrt 2}{3}=\sqrt 2$
$\csc\theta=\dfrac{h}{o}=\dfrac{3\sqrt 2}{3}=\sqrt 2$