## Trigonometry (11th Edition) Clone

$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ The proof is shown below.
$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ We start from the right side. $$X=\frac{2\tan2\theta}{2-\sec^22\theta}$$ - Recall that $\sec^2x=\tan^2x+1$. So if we replace $x$ with $2\theta$, we end up with $\sec^22\theta=\tan^22\theta+1$. Thus, $$X=\frac{2\tan2\theta}{2-(\tan^22\theta+1)}$$ $$X=\frac{2\tan2\theta}{2-\tan^22\theta-1}$$ $$X=\frac{2\tan2\theta}{1-\tan^22\theta}$$ - Now remind yourself that $$\tan2x=\frac{2\tan x}{1-\tan^2x}$$ So again, if we replace $x$ with $2\theta$, the result would be $$\tan(2\times2\theta)=\frac{2\tan2\theta}{1-\tan^22\theta}$$ $$\tan4\theta=\frac{2\tan2\theta}{1-\tan^22\theta}$$ Therefore, $$X=\tan4\theta$$ which means $$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ and the equation is an identity.