How can we show that $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$ 
$$4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}\tag1$$
  $\phi$;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 $(1)={\pi\over 2}$?
 A: according to "How else can we show that"  
Take $f(x)=4\arctan\left({1\over \sqrt{x^3}}\right)-\arctan\left({1\over \sqrt{x^6-1}}\right)$ 
now you can show $f(1.618)=\dfrac{\pi}{2}$

This question is very interesting to me .I was working on $\phi $ properties ,like $\phi^2=\phi+1 \to \phi^3=\phi^2+\phi$ or $\psi=\dfrac{1}{\phi}$ 
we can see this $4\arctan\left({1\over \sqrt{\phi^3}}\right)-\arctan\left({1\over \sqrt{\phi^6-1}}\right)={\pi\over 2}$ as $4a-b=\dfrac{\pi}{2} $ so $4a=\dfrac{\pi}{2} +b \to \tan(4a)=-\cot(b)$  (if you prove this,prove the first relation) 
$$\phi=\dfrac{1+\sqrt5}{2} \\
\phi^3=\phi^2+\phi=(\dfrac{1+\sqrt5}{2})^2+(\dfrac{1+\sqrt5}{2})=\dfrac{1+5+2\sqrt5+2+2\sqrt5}{4}=\dfrac{8+4\sqrt5}{4}=2+\sqrt5 \to \\ 
\phi^6-1=(2+\sqrt5)^2-1=4+5+4\sqrt5-1=8+4\sqrt5$$ now redude to 
$$4\arctan\left({1\over \sqrt{2+\sqrt5}}\right)-\arctan\left({1\over \sqrt{8+4\sqrt5}}\right)=\dfrac{\pi}{2} \to 4a-b=\dfrac{\pi}{2} \to 
\tan(4a)= ? -\cot(b)$$
$$\tan a=\dfrac{1}{\sqrt{2+\sqrt5}} \to 
tan 2a=\dfrac{2\tan a}{1-\tan^2a}=\\\frac{2\dfrac{1}{\sqrt{2+\sqrt5}}}{1-(\dfrac{1}{\sqrt{2+\sqrt5}})^2}=\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5} \\
\tan(4a)=\dfrac{2\tan 2a}{1-\tan^22a}=\\\dfrac{2\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5}}{1-(\dfrac{2\sqrt{2+\sqrt5}}{1+\sqrt5})^2}=\\
\dfrac{4\sqrt{2+\sqrt5}(1+\sqrt5)}{(1+\sqrt5)^2-4(2+\sqrt5)}=\\-2\sqrt{2+\sqrt5}$$ Other side we have $$-\cot b =-(\sqrt{8+4\sqrt5})=-2\sqrt{2+\sqrt5} \checkmark$$
A: A detailed and autonomous prove (without reference to previous publication) :

A: HINT:
You can simplify a lot your equation working the identity
$\phi^2=\phi+1$
Indeed
$\phi^3=\phi^2+\phi=2\phi+1$
and
$\phi^6-1=(2\phi+1)^2-1=4\phi^2+4\phi=4\phi^3$
Therefore your equation becomes
$
4\arctan(a)-\arctan(a/2)=\frac{\pi}{2}
$
with $a=\frac{1}{\sqrt{\phi^3}}$
Using the equality on arctan you provided should be enough to find the solution.
A: Using $\phi^m=F_m\cdot\phi+F_{m-1}$ where $F_n$ is the $n$the Fibonacci Number.
Like Maczinga,
As $\phi^{3/2}>1,$
using showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
$2\arctan\dfrac1{\phi^{3/2}}=\arctan\dfrac{2\phi^{3/2}}{\phi^3-1}=\arctan(\phi^{1/2})$ 
Again $\arctan\dfrac1{\sqrt{\phi^6-1}}=\arctan\dfrac1{2\phi^{3/2}}$
Now, $\arctan\phi^{1/2}-\arctan\dfrac1{2\phi^{3/2}}=\arctan\dfrac{\phi^{1/2}-\dfrac1{2\phi^{3/2}}}{1+\phi^{1/2}\cdot\dfrac1{2\phi^{3/2}}}$ $=\arctan\dfrac{2\phi^2-1}{\phi^{1/2}\cdot(2\phi+1)}$ $=\arctan\dfrac{2(\phi+1)-1}{\phi^{1/2}(2\phi+1)}$
$=\arctan\dfrac1{\phi^{1/2}}$
So, we are left with establishing $\arctan(\phi^{1/2})+\arctan\dfrac1{\phi^{1/2}}=\dfrac\pi2$
