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..

Let $z = \cos 20^\circ\,$ then by the triple angle formula $\frac{1}{2}=\cos 60^\circ = 4 z^3 - 3 z$ $\iff 8z^3-6z-1=0$.
By the double angle formula $\cos 40^\circ = 2 z^2 - 1\,$ and $\cos 80^\circ = 2(2z^2-1)^2-1=8z^4-8z^2+1\,$. But $8z^4=z\cdot8z^3=z(6z+1)$ per the previous equation, so $\cos 80^\circ = -2z^2+z+1\,$.
Then the equality to prove becomes:
$$
\frac{1}{z^2}+\frac{1}{(2z^2-1)^2}+\frac{1}{(2z^2-z-1)^2} = 36
$$
$$
\iff \quad (2z^2-1)^2 (2z^2-z-1)^2 + z^2 (2z^2-z-1)^2 + z^2 (2z^2-1)^2 - 36 z^2(2z^2-1)^2 (2z^2-z-1)^2 = 0
$$
After expanding and routine simplifications, the above reduces to:
$$
576 z^{10} - 576 z^9 - 1024 z^8 + 880 z^7 + 740 z^6 - 452 z^5 - 265 z^4 + 82 z^3 + 41 z^2 - 2 z - 1 = 0
$$
It can be verified by Euclidian division that the latter polynomial has $8z^3-6z-1$ as a factor:
$$
(8 z^3 - 6 z - 1) \cdot (72 z^7 - 72 z^6 - 74 z^5 + 65 z^4 + 28 z^3 - 17 z^2 - 4 z + 1)
$$
Therefore the equality holds, which completes the proof.
