Using De Moivre's Theorm, express $\displaystyle tan 5\theta$ in terms of $\displaystyle tan \theta$.
De Moivre's theorem tells us that:
$\displaystyle (\cos(\theta)+i \sin(\theta))^5=\cos(5\theta)+i \sin(5\theta)$
Now I will write $\displaystyle s$ for $\displaystyle \sin(\theta)$ , $\displaystyle c$ for $\displaystyle \cos(\theta)$ and $\displaystyle t$ for $\displaystyle \tan(\theta)$. Then expanding the left hand side and equating real and imaginary parts we get:
$\displaystyle
\cos(5\theta)=c^5-10c^3s^2+5c~s^4
$
and:
$\displaystyle
\sin(5\theta)=5c^4s-10c^2s^3+s^5
$.
Therefore:
$\displaystyle
\tan(\theta)=\frac{5c^4s-10c^2s^3+s^5}{c^5-10c^3s^2+5c~s^4}
$.
Now on the right divide top and bottom through by $\displaystyle c^5$ and you are done
RonL