Answer
$$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sin x}$$
Work Step by Step
According to Pythagorean Identities,
$$\cot^2x+1=\csc^2 x\hspace{1.5cm}(1)$$
Also, according to Reciprocal Identities,
$$\csc x=\frac{1}{\sin x}$$
So, $$\csc^2 x=\frac{1}{\sin^2 x}\hspace{1.5cm}(2)$$
We can combine $(1)$ and $(2)$ together and have
$$\cot^2 x+1=\frac{1}{\sin^2 x}$$
$$\cot^2 x=\frac{1}{\sin^2 x}-1$$
$$\cot^2 x=\frac{1-\sin^2 x}{\sin^2 x}$$
Now we take the square root of both sides:
$$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sqrt{\sin^2 x}}$$
(Do not forget the $\pm$ sign)
$$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{|\sin x|}$$
Since we already have the $\pm$ sign, we can eliminate the absolute value sign.
$$\cot x=\pm\frac{\sqrt{1-\sin^2 x}}{\sin x}$$
