Question

Prove that, tan3theta - tan2theta - tantheta= tan3theta.tan2theta.tantheta​

Answer 1

Answer:

I am taking theta as a

tan3a - tan2a - tana

As tanb - tanc = (tan (b-c)) (1+tanb tanc)

take b =3a and c = 2a

tan 3a- tan 2a = tan(3a-2a) (1+tan3a tan2a)

= tan a ( 1+ tan 3a tan 2a)

Now, come back to question

tan 3a - tan2a -tan a

= tan a (1+ tan 3a tan2a ) -tan a

=tan a (1+ tan3a tan2a -1)

= tan a tan3a tan 2a

hence proved

#answerwithquality #BAL

Answer 2

Answer:

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