Question

tantheta + tan 2 theta + tan3 theta =tan theta tan2theta tan3 theta​

Answer 1

we have to solve the equation,

tanθ + tan2θ + tan3θ = tanθ.tan2θ.tan3θ

we know,

tan(A + B + C) = (tanA + tanB + tanC - tanA.tanB. tanC)/(1 - tanA.tanB - tanB.tanC - tanC.tanA)

tan(6θ) = tan(θ + 2θ + 3θ) = ( tanθ + tan2θ + tan3θ - tanθ.tan2θ.tan3θ)/(1 - tanθ.tan2θ - tan2θ.tan3θ - tan3θ.tanθ)

here , if we assume , tan(6θ) = 0

then, tanθ + tan2θ + tan3θ = tanθ.tan2θ.tan3θ

hence, tan(6θ) = 0 = tan(π)

or, 6θ = nπ

or, θ = nπ/6 , where n ∈ Z