#### Answer

$$\tan x+\cot x=2\csc2x$$
The equation is an identity, as the left side is equal to the right one.

#### Work Step by Step

$$\tan x+\cot x=2\csc2x$$
The right side would be taken into consideration first.
$$X=2\csc2x$$
We know that $\csc\theta=\frac{1}{\sin\theta}$, so $\csc2x=\frac{1}{\sin2x}$.
$$X=\frac{2}{\sin2x}$$
Now $\sin 2x=2\sin x\cos x$, as stated in Double-Angle Identities.
$$X=\frac{2}{2\sin x\cos x}$$
$$X=\frac{1}{\sin x\cos x}$$
We can rewrite $1$ as $\sin^2x+\cos^2x$, as they also equal $1$.
$$X=\frac{\sin^2x+\cos^2x}{\sin x\cos x}$$
$$X=\frac{\sin^2x}{\sin x\cos x}+\frac{\cos^2x}{\sin x\cos x}$$
$$X=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$$
Applying Quotient Identities: $\frac{\sin x}{\cos x}=\tan x$ and $\frac{\cos x}{\sin x}=\cot x$.
$$X=\tan x+\cot x$$
Thus, $$\tan x+\cot x=2\csc2x$$
The equation is an identity, as the left side is equal to the right one.