For any y , let cot - 1 ( y ) ( 0 , ) and tan - 1 ( y ) ( - 2 , 2 ) . Then the sum of all solutions of the equation tan - 1 ( 6 y 9 - y 2 ) + cot - 1 ( 9 - y 2 6 y ) = 2 3 , for 0 | y | 3 , is equal to [2023]

Question

For any $y \in ℝ$ , let $\cot^{- 1} (y) \in (0, π)$ and $\tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the sum of all solutions of the equation

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3},$ for $0 < | y | < 3$ , is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$Case-I: y \in (- 3, 0) \Rightarrow y < 0 \Rightarrow \frac{6 y}{9 - y^{2}} < 0$

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + π + \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = \frac{2 π}{3}$

$2 \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = - \frac{π}{3}$

$y^{2} - 6 \sqrt{3} y - 9 = 0 \Rightarrow y = 3 \sqrt{3} - 6$ $(∵ y \in (- 3, 0))$

$Case-II: y \in (0, 3) \Rightarrow y > 0 \Rightarrow \frac{6 y}{9 - y^{2}} > 0$

$2 \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = \frac{2 π}{3} \Rightarrow \sqrt{3} y^{2} + 6 y - 9 \sqrt{3} = 0$

$y = \sqrt{3} or y = - 3 \sqrt{3}$ $(rejected)$

$∴$ $Sum = \sqrt{3} + (3 \sqrt{3} - 6) = 4 \sqrt{3} - 6$

Solution

Q.
For any $y \in ℝ$ , let $\cot^{- 1} (y) \in (0, π)$ and $\tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the sum of all solutions of the equation

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3},$ for $0 < | y | < 3$ , is equal to [2023]

1 $2 \sqrt{3} - 3$

2 $3 - 2 \sqrt{3}$

3 $4 \sqrt{3} - 6$

4 $6 - 4 \sqrt{3}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. For any y∈ℝ, let cot-1(y)∈(0,π) and tan-1(y)∈(-π2,π2). Then the sum of all solutions of the equation tan-1(6y9-y2)+cot-1(9-y26y)=2π3, for 0<|y|<3, is equal to [2023]

1 23-3

2 3-23

3 43-6

4 6-43