<s>[INST] Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$ 
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$

I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
 [/INST] Shortest proof:
$$(5+i)^4(239-i)=114244+114244i.$$
Taking the arguments,
$$4\arctan \frac15-\arctan\frac1{239}=\frac\pi4.$$

Note that the computation avoids the fractions and immediately generalizes to other Machin-like formulas (https://en.wikipedia.org/wiki/Machin-like_formula#More_terms).

To perform the computation by hand, consider
$$(5+i)^2=24+10i\propto12+5i,$$
$$(12+5i)^2=119+120i,$$
$$(119+120 i)(239-i)=(119\cdot239+120)+(120\cdot239-119)i\propto 1+i.$$
(After simplification by $119\cdot239$, we have $120=239-119$.)
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