Trigonometry (11th Edition) Clone

$$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2\tan2x$$ The equation is an identity, as shown below.
$$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2\tan2x$$ Let's take on the left side first. $$X=\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}$$ $$X=\frac{(\cos x+\sin x)^2-(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}$$ Recall here that $A^2-B^2=(A-B)(A+B)$, which can be applied to both numerator and denominator in $X$. - Numerator: $$(\cos x+\sin x)^2-(\cos x-\sin x)^2=(\cos x+\sin x-\cos x+\sin x)(\cos x+\sin x+\cos x-\sin x)$$ $$(\cos x+\sin x)^2-(\cos x-\sin x)^2=(2\sin x)\times(2\cos x)$$ $$(\cos x+\sin x)^2-(\cos x-\sin x)^2=4\sin x\cos x$$ $$(\cos x+\sin x)^2-(\cos x-\sin x)^2=2\times(2\sin x\cos x)$$ $$(\cos x+\sin x)^2-(\cos x-\sin x)^2=2\sin 2x$$ (as $\sin2x=2\sin x\cos x$) - Denominator: $$(\cos x-\sin x)(\cos x+\sin x)=\cos^2x-\sin^2x$$ $$(\cos x-\sin x)(\cos x+\sin x)=\cos2x$$ (as $\cos 2x=\cos^2x-\sin^2x$) Combining the numerator and denominator back to $X$, we have $$X=\frac{2\sin2x}{\cos2x}$$ - Quotient Identities: $\frac{\sin2x}{\cos2x}=\tan2x$ $$X=2\tan2x$$ Therefore, $$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2\tan2x$$ We can conclude now that the equation is an identity.