Sum of Solutions of Inverse Trigonometric Equation

To find the sum of all the solutions of the given equation, we analyze the equation: $\tan^{-1}\left(\frac{6y}{9-y^2}\right) + \cot^{-1}\left(\frac{9-y^2}{6y}\right) = \frac{2\pi}{3}$ given that $0 < |y| < 3$, which implies $y \in (-3, 0) \cup (0, 3)$.

Since $0 < |y| < 3$, we have $y^2 < 9$, and therefore $9 - y^2 > 0$.

Let $t = \frac{6y}{9-y^2}$. Since $9 - y^2 > 0$, the sign of $t$ is the same as the sign of $y$.

The given equation simplifies to: $\tan^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right) = \frac{2\pi}{3}$

We now analyze the equation in two cases based on the sign of $y$ (and thus $t$).

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Case 1: $y \in (0, 3)$ (i.e., $t > 0$)

For any $t > 0$, we have: $\cot^{-1}\left(\frac{1}{t}\right) = \tan^{-1}(t)$

Substituting this into the equation gives: $\tan^{-1}(t) + \tan^{-1}(t) = \frac{2\pi}{3}$ $2\tan^{-1}(t) = \frac{2\pi}{3} \implies \tan^{-1}(t) = \frac{\pi}{3}$

Taking the tangent on both sides: $t = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$

Substituting $t = \frac{6y}{9-y^2}$: $\frac{6y}{9-y^2} = \sqrt{3}$ $6y = \sqrt{3}(9 - y^2)$ $\sqrt{3}y^2 + 6y - 9\sqrt{3} = 0$

Dividing the entire equation by $\sqrt{3}$: $y^2 + 2\sqrt{3}y - 9 = 0$

Using the quadratic formula to solve for $y$: $y = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4(1)(-9)}}{2} = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = \frac{-2\sqrt{3} \pm 4\sqrt{3}}{2}$

This gives two roots: $y = \sqrt{3} \quad \text{and} \quad y = -3\sqrt{3}$

Since we restricted $y \in (0, 3)$, the valid solution in this interval is: $y_1 = \sqrt{3}$

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Case 2: $y \in (-3, 0)$ (i.e., $t < 0$)

For $t < 0$, $\frac{1}{t} < 0$. Using the standard definition of $\cot^{-1}(z)$ for $z < 0$: $\cot^{-1}\left(\frac{1}{t}\right) = \pi + \tan^{-1}(t)$

Substituting this into the equation gives: $\tan^{-1}(t) + \pi + \tan^{-1}(t) = \frac{2\pi}{3}$ $2\tan^{-1}(t) = \frac{2\pi}{3} - \pi = -\frac{\pi}{3}$ $\tan^{-1}(t) = -\frac{\pi}{6}$

Taking the tangent on both sides: $t = \tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$

Substituting $t = \frac{6y}{9-y^2}$: $\frac{6y}{9-y^2} = -\frac{1}{\sqrt{3}}$ $-6\sqrt{3}y = 9 - y^2$ $y^2 - 6\sqrt{3}y - 9 = 0$

Using the quadratic formula to solve for $y$: $y = \frac{6\sqrt{3} \pm \sqrt{(-6\sqrt{3})^2 - 4(1)(-9)}}{2} = \frac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = \frac{6\sqrt{3} \pm 12}{2} = 3\sqrt{3} \pm 6$

This gives two roots: $y = 3\sqrt{3} + 6 \approx 11.196 \quad \text{and} \quad y = 3\sqrt{3} - 6 \approx -0.804$

Since we restricted $y \in (-3, 0)$, the valid solution in this interval is: $y_2 = 3\sqrt{3} - 6$

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Sum of Solutions

The sum of all valid solutions is: $\text{Sum} = y_1 + y_2 = \sqrt{3} + (3\sqrt{3} - 6) = 4\sqrt{3} - 6$

Thus, the correct option is (C).