\[tan4 \theta\]
using the identity \[tan2x=\frac{2tanx}{1-tan^2x}\]
(1)Let
\[x=2 \theta\]
then \[tan4 \theta=\frac{2tan 2 \theta}{1-tan^2 \theta}\]
.Now use (1) again with
\[x= \theta\]
.\[\begin{equation} \begin{aligned} tan4 \theta &=\frac{2tan 2 \theta}{1-tan^2 \theta} \\ &= \frac{2(\frac{2tan 2 \theta}{1-tan^2 \theta})}{1-(\frac{2tan 2 \theta}{1-tan^2 \theta})^2} \\ &= \frac{2(\frac{2tan 2 \theta}{1-tan^2 \theta})}{1-(\frac{2tan 2 \theta}{1-tan^2 \theta})^2} \times ( \frac{1-tan^2 \theta}{1-tan^2 \theta} )^2 \\ &= \frac{4 tan \theta (1-tan^2 \theta)}{(1-tan^2 \theta)^2-4 tan^2 \theta} \\ &= \frac{4tan \theta - 4 tan^3 \theta}{1-2 tan \theta + tan^4 \theta -4tan^2 \theta} \\ &= \frac{4tan \theta - 4 tan^3 \theta}{1-6 tan \theta + tan^4 \theta } \end{aligned} \end{equation}\]
.