Answer
$\sin \theta =\dfrac{y}{r}=-\dfrac{\sqrt{21}}{5}$
$\tan \theta =\dfrac{y}{x}=-\dfrac{\sqrt{21}}{2}$
$\csc \theta =\dfrac{1}{\sin \theta}=-\dfrac{5}{\sqrt{21}}=-\dfrac{5\sqrt{21}}{21}$
$\sec \theta =\dfrac{1}{\cos \theta}=\dfrac{5}{2}$
$\cot \theta = \dfrac{1}{\tan \theta}=-\dfrac{2}{\sqrt{21}}=-\dfrac{2\sqrt{21}}{21}$
Work Step by Step
Here, $\cos \theta =\dfrac{2}{5}=\dfrac{x}{r}$
We solve for $y$:
$x^2+y^2=r^2$
$2^2+y^2=5^2$
$y=-\sqrt{21}$
$y$ is negative because we are in quadrant IV (cosine is positive and sine is negative).
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r}=-\dfrac{\sqrt{21}}{5}$
$\tan \theta =\dfrac{y}{x}=-\dfrac{\sqrt{21}}{2}$
$\csc \theta =\dfrac{1}{\sin \theta}=-\dfrac{5}{\sqrt{21}}=-\dfrac{5\sqrt{21}}{21}$
$\sec \theta =\dfrac{1}{\cos \theta}=\dfrac{5}{2}$
$\cot \theta = \dfrac{1}{\tan \theta}=-\dfrac{2}{\sqrt{21}}=-\dfrac{2\sqrt{21}}{21}$
