Consider Machin's Formula
$$4\cot^{-1}{5}-\cot^{-1}{239}={\pi\over 4}\tag1$$
How does one show that
$$\cot^{-1}{\phi}-\cot^{-1}{\phi^3}-\cot^{-1}{\phi^5}-\cot^{-1}{\phi^7}=\cot^{-1}{5}\tag2$$ and $$2\cot^{-1}{\phi}-3\cot^{-1}{\phi^3}-3\cot^{-1}{\phi^5}-4\cot^{-1}{\phi^7}=\cot^{-1}{239}\tag3$$ $\phi$;Golden ratio
An attempt:
$$\cot^{-1}{\phi}-\cot^{-1}{\phi^3}=\cot^{-1}({\phi^2-\phi^{-2}})=\cot^{-1}{3}\tag4$$
$$\cot^{-1}{\phi^5}+\cot^{-1}{\phi^7}=\cot^{-1}{\phi^6-\phi^{-6}\over \sqrt{5}}=\cot^{-1}8\tag5$$
Then $(4)-(5)$
$$\cot^{-1}3-\cot^{-1}8=\cot^{-1}5\tag6$$
$(3)$ seems a bit messy if we apply as above, so I think there is an easy way. How else can we tackle $(3)?$
You may simply consider that $\cot^{-1}(\varphi^k)=\text{arg}(\varphi^k+i)$, hence $(2)$ boils down to checking that
$$ 65 \left(29+13 \sqrt{5}\right)\cdot(\varphi+i) = (5+i)(\varphi^3+i)(\varphi^5+i)(\varphi^7+i) \tag{2}$$ and $(3)$ boils down to checking that: $$ 7140250 \left(375125+167761 \sqrt{5}\right)\cdot(\varphi+i)^2=(239+i)(\varphi^3+i)^3(\varphi^5+i)^3(\varphi^7+i)^4.\tag{3} $$