Answer
$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$
The equation is an identity, as proved below.
Work Step by Step
$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$
From the right side, we examine first.
$$X=\frac{1-\tan^2x}{1+\tan^2x}$$
- Quotient Identity: $\tan x=\frac{\sin x}{\cos x}$
Replace it into $X$:
$$X=\frac{1-\Big(\frac{\sin x}{\cos x}\Big)^2}{1+\Big(\frac{\sin x}{\cos x}\Big)^2}$$
$$X=\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}$$
$$X=\frac{\frac{\cos^2x-\sin^2x}{\cos^2x}}{\frac{\cos^2x+\sin^2x}{\cos^2x}}$$
$$X=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}$$
- Double-Angle Identity: $\cos^2x-\sin^2x=\cos2x$
- Pythagorean Identity: $\cos^2x+\sin^2x=1$
Replace them into $X$:
$$X=\frac{\cos2x}{1}$$
$$X=\cos2x$$
Therefore, $$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$
2 sides are equal, so this must be an identity.
