Spooky Inverse Functions

$\begin{aligned} \sin \left(\tan^{-1} (x) + \color{#3D99F6} \cot^{-1} \left(\frac 1x\right) \right) & = \frac 13 & \small \color{#3D99F6} \text{Note that }\cot^{-1} \left(\frac 1x\right) = \tan^{-1} (x) \\ \sin \left(2 \tan^{-1} (x) \right) & = \frac 13 & \small \color{#3D99F6} \text{By half-angle tangent substitution:} \\ \frac {2x}{1+x^2} & = \frac 13 & \small \color{#3D99F6} \sin \theta = \frac {2 \tan \frac \theta2}{1+\tan^2 \frac \theta 2} \\ x^2 - 6x + 1 & = 0 \\ \implies x & = 6 \pm 2\sqrt 2 \end{aligned}$

Therefore, $a+b+c = 3+2+2 = \boxed 7$ .

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Reference: Half-angle tangent substitution