Trigonometric Sum with a Twist

We prove the following identity: $\frac{1}{x-\tan\theta} + \frac{1}{x-\tan(\theta-\frac{\pi}{3})} + \frac{1}{x-\tan(\theta+\frac{\pi}{3})} = \frac{3x^2-6 x\tan 3\theta - 3}{x^3-3x^2\tan 3\theta-3x+\tan 3\theta}.$ By the triple angle tangent formula: $\tan 3\theta = (3\tan\theta-\tan^3\theta)/(1-3\tan^2\theta)$ , it follows that $x=\tan\theta$ is a zero of the polynomial $P_\theta(x)=x^3-3x^2\tan 3\theta - 3x + \tan 3\theta$ . Moreover, since $\tan 3(\theta + \pi/3) = \tan 3(\theta-\pi/3) = \tan 3\theta$ , it follows that some of the zeros of the polynomial $P_\theta(x)$ include $x=\tan\theta, \tan(\theta+\pi/3),\tan(\theta-\pi/3)$ . Since $P_\theta(x)$ is a monic cubic polynomial, we conclude that these must be all the roots and obtain the following factoring identity: $P_\theta(x) = (x-\tan\theta)\left(x-\tan\left(\theta-\frac{\pi}{3}\right)\right)\left(x-\tan\left(\theta+\frac{\pi}{3}\right)\right).$ The desired identity is obtained from expressing $P'_\theta(x)/P_\theta(x)$ in two different ways: one using the above factoring formula and the other using the unfactored form of $P_\theta(x)$ .

To obtain the value of the expression in question, we substitute $x=\sqrt{3}$ and $\theta = \pi/9 = 20^\circ$ into the main identity. This yields $\frac{1}{\sqrt{3}-\tan(20^\circ)} + \frac{1}{\sqrt{3}+\tan(40^\circ)} + \frac{1}{\sqrt{3}-\tan(80^\circ)} = \frac{\sqrt{3}}{2}.$