Answer
$\sin\theta=\dfrac{2\sqrt{7}}{7}$
$\cos\theta=\dfrac{\sqrt {21}}{7}$
$\tan\theta=\dfrac{2\sqrt 3}{3}$
$\cot\theta=\dfrac{\sqrt 3}{2}$
$\sec\theta=\dfrac{\sqrt{21}}{3}$
$\csc\theta=\dfrac{\sqrt 7}{2}$
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:
$o=2$
$a=\sqrt 3$
Determine the hypotenuse, using the Pythagorean Theorem:
$h^2=o^2+a^2$
$h^2=2^2+(\sqrt 3)^2$
$h^2=7$
$h=\pm\sqrt7$
Since $h$ us never negative, then $h=\sqrt{7}$.
Determine the $6$ trigonometric functions of angle $\theta$:
$\sin\theta=\dfrac{o}{h}=\dfrac{2}{\sqrt 7}=\dfrac{2\sqrt{7}}{7}$
$\cos\theta=\dfrac{a}{h}=\dfrac{\sqrt 3}{\sqrt 7}=\dfrac{\sqrt {21}}{7}$
$\tan\theta=\dfrac{o}{a}=\dfrac{2}{\sqrt 3}=\dfrac{2\sqrt 3}{3}$
$\cot\theta=\dfrac{a}{o}=\dfrac{\sqrt 3}{2}$
$\sec\theta=\dfrac{h}{a}=\dfrac{\sqrt 7}{\sqrt 3}=\dfrac{\sqrt{21}}{3}$
$\csc\theta=\dfrac{h}{o}=\dfrac{\sqrt 7}{2}$
