For any $y \in \mathbb{R}$, let $\cot ^{-1}(y) \in(0, \pi)$ and $\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation
$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0<|y|<3$, is equal to :
$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0<|y|<3$, is equal to :
- A$2 \sqrt{3}-3$
- B$3-2 \sqrt{3}$
- C$4 \sqrt{3}-6$
- D$6-4 \sqrt{3}$
✓ Correct answer: C
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