Trigonometry: Using Double Angle Formula for Tangent to Prove an Equality

**1. The problem statement, all variables and given/known data**

Show that $\displaystyle tan(4arctan(1/5))$ is equivalent to $\displaystyle 120/119$

**2. Relevant equations**

Double Angle Formula for tan:

$\displaystyle tan(2A) =\frac{2tanA}{1-\tan^2A}$

**3. The attempt at a solution**

From my understanding you have to apply the double angle formula twice.

Set x = arctan(1/5) and plug that into the double angle formula to solve for tan(2arctan1/5) (which is basically 2x). Now I set 2x = y, and plug y into the double angle formula to solve for 2y, which is 4arctan1/5.

sub x = arctan(1/5) into the double angle formula:

$\displaystyle tan(2A) =\frac{2tanA}{1-\tan^2A}$

$\displaystyle tan(2arctan(1/5)) =\frac{2tan(arctan1/5)}{1-\tan^2(arctan1/5)}$

Apply tan(arctanx) = x into the formula, resulting with:

$\displaystyle tan(2arctan(1/5)) =\frac{2(1/5)}{1-1/5arctan1/5}$

Simplify the numerator:

$\displaystyle tan(2arctan(1/5)) =\frac{2/5}{1-1/5arctan1/5}$

At this point I'm stuck, did I mess up in simplifying the denominator because I would like to be able to simplify it further so I don't have to use the final result above when I plug it back into a double angle formula once more. If that is correct, how would I work solve for

$\displaystyle tan(2(\frac{2/5}{1-1/5arctan1/5})) =\frac{2tan(\frac{2/5}{1-1/5arctan1/5})}{1-\tan^2(\frac{2/5}{1-1/5arctan1/5})}$?

Thanks for any assistance. :)