Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$ Without any software and approximations prove that $$\sec(52^{\circ})-\cos(52^{\circ})>1$$
We can use some known trig values like $18^{\circ}$,$54^{\circ}$,etc
My try:
I considered the function:
$$f(x)=\sec(x)-\cos(x)-1,\: x\in \left (0, \frac{\pi}{3}\right)$$
We have the derivative as:
$$f'(x)=\sec x\tan x+\sin x >0$$ so $f$ is Monotone increasing.
So we have:
$$f(52^{\circ})>f(45^{\circ})=\frac{1}{\sqrt{2}}-1$$ but not able to proceed

I propose another way using the idea of golden ratio with triple-angle formula.
We have that for $\theta\in(0,\pi/2)$, $f(\theta)=\sec \theta -\cos \theta$ is an increasing function and
$$\sec \theta-\cos \theta =1 \implies \cos \theta = \frac{\sqrt 5-1}2 =\frac 1\varphi$$
Since $\theta < \beta =54°$ we have that
$$\theta < \alpha =52° \iff \cos \theta<\cos \alpha \iff \cos (3\theta)> \cos (3\alpha)$$
with
$$\cos(3\theta)= 4\cos^3 \theta-3\cos \theta = \frac4{\varphi^3}-\frac3{\varphi}=5\varphi-9$$
and using golden gnomon properties

$$\cos(3\alpha)=\cos( 4\beta-60°)=\cos(4\beta)\cos(60°)+\sin(4\beta)\sin(60°)=$$
$$=-\cos(36°)\cos(60°)-\sin(36°)\sin(60°)=-\frac{\varphi}4-\frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}$$
which leads to
$$\cos (3\theta)> \cos (3\alpha)\iff \frac{\sqrt 3}2\sqrt{\frac 3 4 -\frac \varphi 4}>9-\frac{21}4 \varphi \iff \varphi >\frac{144}{89}$$
which is true, indeed $F_{11}=89$ and $F_{12}=144$ are two consecutive Fibonacci numbers and their ratios approximate $\varphi$ such that
$$\frac{F_{n+1}}{F_{n}}<\varphi $$
for $n$ odd, therefore $\theta <52°$ and the given inequality holds.
