Answer
$\sin\theta=\dfrac{\sqrt{3}}{2}$
$\cos\theta=\dfrac{1}{2}$
$\tan\theta=\sqrt 3$
$\cot\theta=\dfrac{\sqrt 3}{3}$
$\sec\theta=2$
$\csc\theta=\dfrac{2\sqrt 3}{3}$
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:
$h=4$
$a=2$
Determine the opposite side, using the Pythagorean Theorem:
$h^2=o^2+a^2$
$4^2=o^2+2^2$
$o^2=12$
$o=\pm\sqrt{12}$
$o=\pm2\sqrt{3}$
Since $h$ is never negative, then $h=2\sqrt3$.
Determine the 6 trigonometric functions of angle $\theta$:
$\sin\theta=\dfrac{o}{h}=\dfrac{2\sqrt 3}{4}=\dfrac{\sqrt{3}}{2}$
$\cos\theta=\dfrac{a}{h}=\dfrac{2}{4}=\dfrac{1}{2}$
$\tan\theta=\dfrac{o}{a}=\dfrac{2\sqrt 3}{2}=\sqrt 3$
$\cot\theta=\dfrac{a}{o}=\dfrac{2}{2\sqrt 3}=\dfrac{1}{\sqrt 3}=\dfrac{\sqrt 3}{3}$
$\sec\theta=\dfrac{h}{a}=\dfrac{4}{2}=2$
$\csc\theta=\dfrac{h}{o}=\dfrac{4}{2\sqrt 3}=\dfrac{2}{\sqrt 3}=\dfrac{2\sqrt 3}{3}$
