TITLE: How can we show that $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$
QUESTION [6 upvotes]: $$4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}\tag1$$
  $\phi$;Golden ratio

I understand that we can use 
$$\arctan{1\over a}+\arctan{1\over b}=\arctan{a+b\over ab-1}$$
that would take quite a long time and simplifying algebraic expressions involving surds is also difficult task to do.
How else can we show that $(1)={\pi\over 2}$?

REPLY [2 votes]: A detailed and autonomous prove (without reference to previous publication) :