Answer
$$\cot x=-\frac{4}{5}$$
$$\sec x=-\frac{\sqrt{41}}{4}$$
$$\csc x=\frac{\sqrt{41}}{5}$$
Work Step by Step
$$\tan x=-\frac{5}{4}\hspace{2cm}\frac{\pi}{2}\lt x\lt \pi$$
1) Determining the signs of $\cot x$, $\sec x$ and $\csc x$
The place of $x$ is where $\frac{\pi}{2}\lt x\lt \pi$, meaning $x$ is in quadrant II, where sines are positive but cosines are negative.
As $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$, this means both tangents and cotangents in quadrant II are negative.
As $\sec x=\frac{1}{\cos x}$ and $\csc x=\frac{1}{\sin x}$, this means secants are negative, while cosecants are positive in quadrant II.
Therefore, $\cot x\lt0$, $\sec x\lt0$ and $\csc x\gt0$.
2) Calculating $\cot x$
- Recall the Reciprocal Identities: $\cot x=\frac{1}{\tan x}$
Therefore, $$\cot x=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}$$
3) Calculating $\sec x$ and $\csc x$
- Recall the Pythagorean Identities for secant and cosecants: $$\sec^2x=1+\tan^2x$$
$$\csc^2x=1+\cot^2x$$
Therefore,
$$\sec^2x=1+\Big(-\frac{5}{4}\Big)^2=1+\frac{25}{16}=\frac{41}{16}$$
$$\sec x=\pm\frac{\sqrt{41}}{4}$$
$$\csc^2x=1+\Big(-\frac{4}{5}\Big)^2=1+\frac{16}{25}=\frac{41}{25}$$
$$\csc x=\pm\frac{\sqrt{41}}{5}$$
Since $\sec x\lt0$, $$\sec x=-\frac{\sqrt{41}}{4}$$
Since $\csc x\gt0$, $$\csc x=\frac{\sqrt{41}}{5}$$
