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Mathematics Question on Methods of Integration

For any y∈R, let cot−1^{-1}(y) ∈ (0,π) and tan−1^{-1}(y) ∈ (−π2,π2-\frac{\pi}{2},\frac{\pi}{2}). Then the sum of all the solutions of the equationtan−1(6y9−y2)+cot−1(9−y26y)=2π3 for 0<∣y∣<3,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

23−32\sqrt3-3

B

3−233-2\sqrt3

C

43−64\sqrt3-6

D

6−436-4\sqrt3

Answer

43−64\sqrt3-6

Explanation

Solution

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