Answer
$\sin\theta=\dfrac{2\sqrt{13}}{13}$
$\cos\theta=\dfrac{3\sqrt{13}}{13}$
$\tan\theta=\dfrac{2}{3}$
$\cot\theta=\dfrac{3}{2}$
$\sec\theta=\dfrac{\sqrt{13}}{3}$
$\csc\theta=\dfrac{\sqrt{13}}{2}$
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:
$o=2$
$a=3$
Determine the hypotenuse, using the Pythagorean Theorem:
$h^2=o^2+a^2$
$h^2=2^2+3^2$
$h^2=13$
$h=\pm\sqrt{13}$
Since $h$ is never negative, then $h=\sqrt{13}$.
Determine the $6$ trigonometric functions of angle $\theta$:
$\sin\theta=\dfrac{o}{h}=\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}$
$\cos\theta=\dfrac{a}{h}=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}$
$\tan\theta=\dfrac{o}{a}=\dfrac{2}{3}$
$\cot\theta=\dfrac{a}{o}=\dfrac{3}{2}$
$\sec\theta=\dfrac{h}{a}=\dfrac{\sqrt{13}}{3}$
$\csc\theta=\dfrac{h}{o}=\dfrac{\sqrt{13}}{2}$
