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

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Inverse Trigonometric Functions

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 the 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

Answer: (C)

Case I:

y \in (- 3, 0)

For negative

y

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

So:

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

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

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

\frac{6 y}{9 - y ^{2}} = - \frac{1}{3}

63 y = - (9 - y^{2})

y^{2} - 63 y - 9 = 0

y = 33 - 6

(taking the value in

(- 3, 0)

)

Case II:

y \in (0, 3)

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

\frac{6 y}{9 - y ^{2}} = 3

3 y^{2} + 6 y - 93 = 0

y = 3

(taking the positive root)

Sum =

3 + 33 - 6 = 43 - 6

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Single correct option