#### Answer

$$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$
The equation is an identity as proved below.

#### Work Step by Step

$$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$
The left side would be analyzed first.
$$X=\csc^4x-\cot^4x$$
$$X=(\csc^2x-\cot^2x)(\csc^2x+\cot^2x)$$
- $\csc x$ and $\cot x$ would be rewritten according to the following identities:
$$\csc x=\frac{1}{\sin x}\hspace{2cm}\cot x=\frac{\cos x}{\sin x}$$
Therefore, $$\csc^2x=\frac{1}{\sin^2x}\hspace{2cm}\cot^2x=\frac{\cos^2x}{\sin^2x}$$
Apply them into $X$:
$$X=\Big(\frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}\Big)\Big(\frac{1}{\sin^2x}+\frac{\cos^2x}{\sin^2x}\Big)$$
$$X=\frac{1-\cos^2x}{\sin^2x}\times\frac{1+\cos^2x}{\sin^2x}$$
$$X=\frac{(1-\cos^2x)(1+\cos^2x)}{\sin^4x}$$
- We can rewrite $(1-\cos^2x)$ on the numerator into $\sin^2x$ according to Pythagorean identity.
$$X=\frac{\sin^2x(1+\cos^2x)}{\sin^4x}$$
$$X=\frac{1+\cos^2x}{\sin^2x}$$
- Finally, also from Pythagorean identity, $\sin^2x$ in the denominator can be written into $(1-\cos^2x)$
$$X=\frac{1+\cos^2x}{1-\cos^2x}$$
So we can conclude that $$\csc^4x-\cot^4x=\frac{1+\cos^2x}{1-\cos^2x}$$
The equation is therefore an identity.