Answer
$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$
The proof is shown below.
Work Step by Step
$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$
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.
