<s>[INST] Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\cos^2 40°}{\cos^2 20°.\cos^2 40°.\cos^2 80°}$$.
I got paused here. Please help to prove this..
 [/INST] Starting like dxiv,
let $a=\cos20^\circ,b=-\cos40^\circ,c=-\cos80^\circ,$
As $\cos(3\cdot20^\circ)=\dfrac12$
$\cos(3\cdot40^\circ)=-\dfrac12$
$\cos(3\cdot80^\circ)=-\dfrac12$
As $\cos3x=\cos60^\circ, 3x=360^\circ m\pm60^\circ$ where $m$ is any integer.
$\implies x=120^\circ m+20^\circ$ where $m\equiv-1,0,1\pmod3$
Now as $\cos3x=4\cos^3x-3\cos x$
The roots of $4t^3-3t-\dfrac12=0$ are $a,b,c$
and we need to find $\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}=\dfrac{a^2b^2+b^2c^2+c^2a^2}{(abc)^2}$
$=\dfrac{(ab+bc+ca)^2-2abc(a+b+c)}{(abc)^2}$
By Vieta's formula 
$a+b+c=\dfrac04$
$ab+bc+ca=-\dfrac34$
$abc=-\dfrac1{2\cdot4}$
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