## Trigonometry (11th Edition) Clone

$$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$ The detailed step of verification is in the Work step by step.
$$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$ We start from the left side one more time. $$X=\frac{\cos^4x-\sin^4x}{\cos^2x}$$ $$X=\frac{(\cos^2x)^2-(\sin^2x)^2}{\cos^2x}$$ $$X=\frac{(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)}{\cos^2x}$$ ($A^2-B^2=(A-B)(A+B)$) - $\cos^2x+\sin^2x$ is equal to $1$, so we write it into $1$. $$X=\frac{(\cos^2x-\sin^2x)\times1}{\cos^2x}$$ $$X=\frac{\cos^2x-\sin^2x}{\cos^2x}$$ $$X=1-\frac{\sin^2x}{\cos^2x}$$ - As $\tan x=\frac{\sin x}{\cos x}$, it follows that $\tan^2x=\frac{\sin^2x}{\cos^2x}$ $$X=1-\tan^2x$$ So, $$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$ The equation is an identity as a result.