Proving $\cos 36° > \tan 36° $ How do we  prove that  $\cos  36° > \tan 36° $ ? Please help . Thank you.

Since $\cos x$ and $\tan x$ are monotonic in the $[0,\pi/2]$, they cross at a single point, namely when:
$$\cos^2x=\sin x\ \rightarrow\  1 -\sin^2x-\sin x =0 $$
Or when:
$$\sin x=\varphi-1\ \rightarrow \ x\approx38.2^\circ$$
Since $36^\circ<38.2^\circ$, we also have $\cos 36^\circ >\tan 36^\circ$.
$_\blacksquare$
