TITLE: How to prove that $4 \cot ^{-1}\left(\sqrt{\phi }\right)+\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)=\pi$
QUESTION [6 upvotes]: Trying to answer this question (five years too late), I found to the surprising identity
$$4 \cot ^{-1}\left(\sqrt{\phi }\right)+\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17
   \sqrt{5}}\right)=\pi$$
How to prove it ?

REPLY [6 votes]: Use the formula
$$\cot2\alpha=\frac{\cot^2\alpha-1}{2\cot\alpha}$$
twice to conclude that
$$
\cot(4\mathrm{arccot}\sqrt{\phi})=-\frac18\sqrt{\phi}-\frac58\sqrt{5\phi}.
$$
The quantity on the right is equal to $-\dfrac14\sqrt{22+17\sqrt5}$. Together with $\cot(\pi-\alpha)=-\cot\alpha$ this leads to an answer. Manipulations of nested square roots are left to the reader.