Question

For any y∈R, let cot\(^{-1}\)(y) ∈ (0,π) and tan\(^{-1}\)(y) ∈ (\(-\frac{\pi}{2},\frac{\pi}{2}\)). Then the sum of all the solutions of the equation\(tan^{-1}(\frac{6y}{9-y^2})+cot^{-1}(\frac{9-y^2}{6y})=\frac{2\pi}{3}\, for\,0<|y|<3,\) is equal to

• A. \(2\sqrt3-3\)
• B. \(3-2\sqrt3\)
• C. \(4\sqrt3-6\)
• D. \(6-4\sqrt3\)

Answer 1

The Correct Option is C

The solutions combine to yield the sum of solutions, which is calculated as √3 + (3√3 - 6).

The correct option is (C): 4√3-6

Answer 2

Given,

\(\tan^{-1}\left(\frac{6y}{9-y^2}\right) + \cot^{-1}\left(\frac{9-y^2}{6y}\right) = \frac{2\pi}{3} \quad 0 < |y| < 3\)

\(\Rightarrow y \in (-3, 3) - \{0\}\)

breaks down the problem into two cases based on the sign of \(\frac{6y}{9-y^2}\)​:

Case I: When \(\frac{6y}{9-y^2} \gt 0\) (which implies \(y\gt 0\))

Case II: When \(\frac{6y}{9-y^2} \lt 0\) (which implies \(y\lt0\))

Case I:
\(\frac{6y}{9-y^2} > 0 \Rightarrow y > 0\)

\(\tan^{-1}\left(\frac{6y}{9-y^2}\right) + \tan^{-1}\left(\frac{6y}{9-y^2}\right) = \frac{2\pi}{3}\)

\(\Rightarrow 2 \tan^{-1}\left(\frac{6y}{9-y^2}\right) = \frac{2\pi}{3}\)

\(\Rightarrow \tan^{-1}\left(\frac{6y}{9-y^2}\right) = \frac{\pi}{3} \Rightarrow \frac{6y}{9-y^2} = \sqrt{3}\)

\(\Rightarrow 6y = 9\sqrt{3} - \sqrt{3}y^2\)
Solving the Quadratic Equation:
\(\Rightarrow \sqrt{3}y^2 + 6y - 9\sqrt{3} = 0\)
\(\Rightarrow \sqrt{3}y^2 + 9y - 3y - 9\sqrt{3} = 0\)
\(\Rightarrow \sqrt{3}y(y + 3\sqrt{3}) - 3(y + 3\sqrt{3}) = 0\)
\(\Rightarrow (y + 3\sqrt{3})(\sqrt{3}y - 3) = 0\)
\(y \neq -3\sqrt{3}\)
\(y=\sqrt3\ as\ y\in(0,3)\)

Case II:

\(\frac{6y}{9 - y^2} < 0 \Rightarrow y < 0\)

\(\tan^{-1} \left( \frac{6y}{9 - y^2} \right) + \pi + \tan^{-1} \left( \frac{6y}{9 - y^2} \right) = \frac{2\pi}{3}\)

\(\Rightarrow 2 \tan^{-1} \left( \frac{6y}{9 - y^2} \right) + \pi = \frac{2\pi}{3} \Rightarrow 2 \tan^{-1} \left( \frac{6y}{9 - y^2} \right) = -\frac{\pi}{3} \Rightarrow \tan^{-1} \left( \frac{6y}{9 - y^2} \right) = -\frac{\pi}{6}\)

\(\Rightarrow \frac{6y}{9 - y^2} = -\frac{1}{\sqrt{3}}\)

\(\Rightarrow 6y\sqrt{3} = -9 + y^2\)
Solving the Quadratic Equation:
\(\Rightarrow y^2 - 6\sqrt{3}y - 9 = 0\)

\(\Rightarrow y = \frac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = \frac{6\sqrt{3} \pm 12}{2} = 3\sqrt{3} \pm 6\)

\(\text{as } y \in (-3, 0) \quad \therefore y = 3\sqrt{3} - 6\)
Now add both answers
\(y = \sqrt{3} +(3\sqrt{3} - 6)\)
\(y = 4\sqrt{3} - 6\)

So, the correct option is (C): \(4\sqrt{3} - 6\)