#### Answer

$$\sin x=-\frac{4}{5}$$
$$\tan x=-\frac{3}{4}$$
$$\cot(-x)=\frac{4}{3}$$

#### Work Step by Step

$$\cos x=\frac{3}{5}\hspace{2cm}\text{x in quadrant IV}$$
1) Determining the signs of $\sin x$, $\tan x$ and $\cot x$
$x$ is in quadrant IV, where sines are negative but cosines are positive.
As, $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$, this means both tangents and cotangents in quadrant are negative.
Therefore, $\sin x\lt0$, $\tan x\lt0$ and $\cot x\lt0$.
2) Calculating $\sin x$
- Recall the Pythagorean Identities: $\sin^2x=1-\cos^2x$
Therefore, $$\sin^2x=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$
$$\sin x=\pm\frac{4}{5}$$
But as $\sin x\lt0$, $$\sin x=-\frac{4}{5}$$
3) Calculating $\tan x$
- Recall the Quotient Identities: $$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$$
4) Calculating $\cot(-x)$
- Recall the Quotient Identities:
$$\cot x=\frac{\cos x}{\sin x}=\frac{-\frac{4}{5}}{\frac{3}{5}}=-\frac{4}{3}$$
- However, also recall the Negative-Angle Identities for cotangents:
$$\cot(-x)=-\cot x=-\Big(-\frac{4}{3}\Big)=\frac{4}{3}$$