#### Answer

$$1+\tan x\tan2x=\sec2x$$
The equation is an identity since two sides are equal to each other as shown below.

#### Work Step by Step

$$1+\tan x\tan2x=\sec2x$$
We started dealing with the left side.
$$X=1+\tan x\tan2x$$
- From Double-Angle Identity for tangent, we get
$$\tan2x=\frac{2\tan x}{1-\tan^2x}$$
Thus, replace into $X$:
$$X=1+\tan x\times\frac{2\tan x}{1-\tan^2x}$$
$$X=1+\frac{2\tan^2x}{1-\tan^2x}$$
Now, the right side has $\sec2x$, which is related to $\cos2x$, we probably should write $\tan x$ as $\frac{\sin x}{\cos x}$ here.
$$X=1+\frac{2\times\frac{\sin^2x}{\cos^2x}}{1-\frac{\sin^2x}{\cos^2x}}$$
$$X=1+\frac{\frac{2\sin^2x}{\cos^2x}}{\frac{\cos^2x-\sin^2x}{\cos^2x}}$$
2 $\cos^2x$ would eliminate each other.
$$X=1+\frac{2\sin^2x}{\cos^2x-\sin^2x}$$
$$X=\frac{\cos^2x-\sin^2x+2\sin^2x}{\cos^2x-\sin^2x}$$
$$X=\frac{\cos^2x+\sin^2x}{\cos^2x-\sin^2x}$$
- From Pythagorean Identities: $\cos^2x+\sin^2x=1$
- From Double-Angle Identities for cosine: $\cos^2x-\sin^2x=\cos2x$
$$X=\frac{1}{\cos2x}$$
$$X=\sec2x\hspace{1cm}\text{(Reciprocal Identity)}$$
Therefore, $$1+\tan x\tan2x=\sec2x$$
We can conclude that the equation is an identity.