If y∈(−3,0),tan−1(6y9−y2)+cot−1(9−y26y)=2π3 then 33−y=____

Tan−1(1x)=−π+cot−1(x),x<0 Tan−1(6y9−y2)+π+Tan−1(6y9−y2)=2π3 2Tan−1(6y9−y2)=−π3 (6y9−y2)=−13 y2−63y−9=0 y=33−6

If y∈(−3,0),tan−1(6y9−y2)+cot−1(9−y26y)=2π3 then 33−y=____

Tan−1(1x)=−π+cot−1(x),x<0 Tan−1(6y9−y2)+π+Tan−1(6y9−y2)=2π3 2Tan−1(6y9−y2)=−π3 (6y9−y2)=−13 y2−63y−9=0 y=33−6

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If $y \in (- 3, 0), \tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3}$ then $3 \sqrt{3} - y$ =____

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answer is C.

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Detailed Solution

$T a n^{- 1} (\frac{1}{x}) = - π + \cot^{- 1} (x), x < 0 T a n^{- 1} (\frac{6 y}{9 - y^{2}}) + π + T a n^{- 1} (\frac{6 y}{9 - y^{2}}) = \frac{2 π}{3} 2 T a n^{- 1} (\frac{6 y}{9 - y^{2}}) = - \frac{π}{3} (\frac{6 y}{9 - y^{2}}) = - \frac{1}{\sqrt{3}} y^{2} - 6 \sqrt{3} y - 9 = 0 y = 3 \sqrt{3} - 6$