Trigonometry (11th Edition) Clone

$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$ The equation is an identity, as proved below.
$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$ 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.