Answer
$\sin\theta=\dfrac{2\sqrt{13}}{13}$
$\cos\theta=\dfrac{3\sqrt{13}}{13}$
$\tan\theta=\dfrac{2}{3}$
$\csc\theta=\dfrac{\sqrt{13}}{2}$
$\sec\theta=\dfrac{\sqrt{13}}{3}$
$\cot\theta=\dfrac{3}{2}$
Work Step by Step
The triangle is shown in the attached image below.
Let $h$ be the hypotenuse of the triangle. Use the Pythagorean Theorem to find it:
$h=\sqrt{3^{2}+2^{2}}=\sqrt{9+4}=\sqrt{13}$
$\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}$
$\cos\theta=\dfrac{adjacent}{hypotenuse}=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}$
$\tan\theta=\dfrac{opposite}{adjacent}=\dfrac{2}{3}$
$\csc\theta=\dfrac{hypotenuse}{opposite}=\dfrac{\sqrt{13}}{2}$
$\sec\theta=\dfrac{hypotenuse}{adjacent}=\dfrac{\sqrt{13}}{3}$
$\cot\theta=\dfrac{adjacent}{opposite}=\dfrac{3}{2}$