Answer
$tan~4\theta = \frac{2~tan~2\theta}{2-sec^2~2\theta}$
Work Step by Step
We can verify that the equation is an identity:
$tan~4\theta$
$= \frac{sin~4\theta}{cos~4\theta}$
$= \frac{2~sin~2\theta~cos~2\theta}{2~cos^2~2\theta-1}$
$= \frac{2~sin~2\theta~cos~2\theta}{2~cos^2~2\theta-1}\cdot \frac{\frac{1}{cos^2~2\theta}}{\frac{1}{cos^2~2\theta}}$
$= \frac{\frac{2~sin~2\theta~cos~2\theta}{cos^2~2\theta}}{\frac{2~cos^2~2\theta-1}{cos^2~2\theta}}$
$= \frac{\frac{2~sin~2\theta}{cos~2\theta}}{2-\frac{1}{cos^2~2\theta}}$
$= \frac{2~tan~2\theta}{2-sec^2~2\theta}$
