If tan⁻¹ x + tan⁻¹ y + tan⁻¹ z = π, then prove that x + y + z = xyz.
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Solution :
i ) Let tan^-1 x = A
=> x = tanA
ii ) tan^-1 y = B
=> y = tan B
iii ) tan^-1 z = C
=> z = tanC
Given A + B + C = π
=> A + B = π - C
=> tan( A + B ) = tan( π - C )
=> ( tanA + tanB )/(1-tanAtanB ) = -tanC
=> tanA+tanB = -tanC + tanAtanBtanC
=> tanA+tanB+tanC = tanAtanBtanC
=> x + y + z = xyz
i ) Let tan^-1 x = A
=> x = tanA
ii ) tan^-1 y = B
=> y = tan B
iii ) tan^-1 z = C
=> z = tanC
Given A + B + C = π
=> A + B = π - C
=> tan( A + B ) = tan( π - C )
=> ( tanA + tanB )/(1-tanAtanB ) = -tanC
=> tanA+tanB = -tanC + tanAtanBtanC
=> tanA+tanB+tanC = tanAtanBtanC
=> x + y + z = xyz
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