A trigonometric identity: $\frac1{\sin40 ^\circ}+\tan10^\circ=\sqrt{3}.$ My sister asked me such a trigonometric identity (her high school challenging problem):

prove:

$$\frac1{\sin40 ^\circ}+\tan10^\circ=\sqrt{3}.$$
I found that this is really true (surprising... with a calculator), but as an undergraduate equipped with calculus and linear algebra, I have no idea how to attack this problem. Thanks in advance for any help!

Replace $\sqrt{3}$ with $2 \sin{60^\circ}$. We want to show $$\frac1{\sin40 ^\circ}+\tan10^\circ=2 \sin{60^\circ}.$$
Let $q = e^{2 i \pi / 360}$ (A nome of one degree) then ​$$\sin(\theta^\circ) = \frac{q^\theta - q^{-\theta}}{2i}$$ and ​$$\tan(\theta^\circ) = - i \frac{q^{\theta} - q^{-\theta}}{q^{\theta} + q^{-\theta}}$$ so we can rewrite our equation in terms of the nome:
$$\frac{2i}{q^{40} - q^{-40}} - i \frac{q^{10} - q^{-10}}{q^{10} + q^{-10}} - 2 \frac{q^{60} - q^{-60}}{2i} = 0$$
multiplying this out and then reducing it modulo the cyclotomic polynomial $\Phi_{360}(q) = q^{96} + q^{84} - q^{60} - q^{48} - q^{36} + q^{12} + 1$ gives zero:
? r = (2*I)/(q^40 - q^(-40)) - I*(q^10-q^(-10))/(q^10+q^(-10)) - 2*(q^60-q^(-60))/(2*I)
% = (-q^200 + q^140 - q^120 - q^80 + q^60 - 1)/(I*q^140 - I*q^60)
? Mod(r, q^96 + q^84 - q^60 - q^48 - q^36 + q^12 + 1)
% = Mod(0, q^96 + q^84 - q^60 - q^48 - q^36 + q^12 + 1)

this proves the identity.
