Answer
$$\sin2x=\frac{2\tan x}{1+\tan^2x}$$
The equation is an identity.
Work Step by Step
$$\sin2x=\frac{2\tan x}{1+\tan^2x}$$
The right side is more complex, so we need to deal with it first.
$$X=\frac{2\tan x}{1+\tan^2x}$$
- From Quotient Identity: $$\tan x=\frac{\sin x}{\cos x}$$
Therefore, $$X=\frac{\frac{2\sin x}{\cos x}}{1+\frac{\sin^2x}{\cos^2x}}$$
$$X=\frac{\frac{2\sin x}{\cos x}}{\frac{\cos^2x+\sin^2x}{\cos^2x}}$$
$$X=\frac{(2\sin x)\cos^2x}{(\cos^2x+\sin^2x)\cos x}$$
$$X=\frac{2\sin x\cos x}{\cos^2x+\sin^2x}$$
- From Pythagorean Identity: $\cos^2x+\sin^2x=1$
- From Double-Angle Identity: $2\sin x\cos x=\sin2x$
$$X=\frac{\sin 2x}{1}$$
$$X=\sin2x$$
So, $$\sin2x=\frac{2\tan x}{1+\tan^2x}$$
The equation is an identity.