Answer
$$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2\tan2x$$
The equation is an identity, as shown below.
Work Step by Step
$$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2\tan2x$$
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.
