For any y in R,let cot ^{-1}(y) in(0, pi) and | vedclass

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Question

(C) Given equation: $	an ^{-1}\left(\frac{6 y}{9-y^2}ight)+\cot ^{-1}\left(\frac{9-y^2}{6 y}ight)=\frac{2 \pi}{3}$.
Let $x = \frac{6y}{9-y^2}$.

Vedclass · Accepted Answer

$4 \sqrt{3}-6$

Vedclass · Answer

(C) Given equation: $ an ^{-1}\left(\frac{6 y}{9-y^2} ight)+\cot ^{-1}\left(\frac{9-y^2}{6 y} ight)=\frac{2 \pi}{3}$. Let $x = \frac{6y}{9-y^2}$. Then the equation is $ an^{-1}(x) + \cot^{-1}(\frac{1}{x}) = \frac{2\pi}{3}$. Case $I$: If $x > 0$,then $\cot^{-1}(\frac{1}{x}) = an^{-1}(x)$. The equation becomes $2 an^{-1}(x) = \frac{2\pi}{3} \Rightarrow an^{-1}(x) = \frac{\pi}{3} \Rightarrow x = \sqrt{3}$. $\frac{6y}{9-y^2} = \sqrt{3} \Rightarrow 6y = 9\sqrt{3} - \sqrt{3}y^2 \Rightarrow \sqrt{3}y^2 + 6y - 9\sqrt{3} = 0 \Rightarrow y^2 + 2\sqrt{3}y - 9 = 0$. Solving for $y$: $y = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = -\sqrt{3} \pm \sqrt{12} = -\sqrt{3} \pm 2\sqrt{3}$. Since $x > 0$,we need $y \in (0, 3)$,so $y = \sqrt{3}$. Case $II$: If $x < 0$,then $\cot^{-1}(\frac{1}{x}) = an^{-1}(x) + \pi$. The equation becomes $2 an^{-1}(x) + \pi = \frac{2\pi}{3} \Rightarrow 2 an^{-1}(x) = -\frac{\pi}{3} \Rightarrow an^{-1}(x) = -\frac{\pi}{6} \Rightarrow x = -\frac{1}{\sqrt{3}}$. $\frac{6y}{9-y^2} = -\frac{1}{\sqrt{3}} \Rightarrow 6\sqrt{3}y = -9 + y^2 \Rightarrow y^2 - 6\sqrt{3}y - 9 = 0$. Solving for $y$: $y = \frac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = 3\sqrt{3} \pm \sqrt{36} = 3\sqrt{3} \pm 6$. Since $x < 0$,we need $y \in (-3, 0)$,so $y = 3\sqrt{3} - 6$. Sum of solutions: $\sqrt{3} + 3\sqrt{3} - 6 = 4\sqrt{3} - 6$.

For any $y \in R$,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

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