#### Answer

$$\csc x=\pm\frac{\sqrt{1-\cos^2 x}}{1-\cos^2 x}$$

#### Work Step by Step

From Reciprocal Identities:
$$\csc x=\frac{1}{\sin x}\hspace{1.5cm}(1)$$
We see that we find a connection between $\csc x$ and $\sin x$. However, the exercise asks for a connection between $\csc x$ and $\cos x$. What we can do is find a way to represent $\sin x$ in terms of $\cos x$
From Pythagorean Identities, we also have:
$$\sin^2 x+\cos^2 x=1$$
which means $$\sin^2 x=1-\cos^2 x$$
$$\sin x=\pm\sqrt{1-\cos^2 x}\hspace{1.5cm}(2)$$ (as we take the square root of both sides)
As we found the way to represent $\sin x$ in terms of $\cos x$, we can adopt it for $\csc x$ in $(1)$:
$$\csc x=\frac{1}{\pm\sqrt{1-\cos^2 x}}$$
$$\csc x=\pm\frac{1}{\sqrt{1-\cos^2 x}}$$
We rationalize the denominator by multiplying both numerator and denominator by $\sqrt{1-\cos^2 x}$
$$\csc x=\pm\frac{1}{\sqrt{1-\cos^2 x}}\frac{\sqrt{1-\cos^2 x}}{\sqrt{1-\cos^2 x}}$$
$$\csc x=\pm\frac{\sqrt{1-\cos^2 x}}{1-\cos^2 x}$$