Q3/075) Evaluate tan^2 20 + tan^2 40 + tan^2 80.

Observe that tan(3⋅20∘)=tan60=√3

-  tan(3⋅40)=- tan120=- tan(180−60)=tan60=√3  

and tan(3⋅80)=tan 240=tan(180+60)=tan60=√3

As tan3θ=(3tanθ−tan^3θ)/)1−3tan^2 θ)

the roots of the equation t^3−3√3t^2−3t+√3=0 ( Putting tan3θ=√3)

will be tan20,tan(−40)=−tan40∘,tan80∘

that is of f(t) =t^3−3√3t^2−3t+√3=0
and we shall construct an equation whose roots are
tan^2 20,tan^2 40 tan^2 80
shall be f(x^(1/2) = 0)
putting t = x^(1/2) we get
so x^(3/2) - 3√3x−3x^(1/2) +√3=0
or x^(3/2) - 3 x^(1/2) = 3√3x -√3
or
√x(x-3) = √3(3x -1)
square both sides to get
x(x-3)^3 = 3(3x-1)^2
or x(x^2-6x+ 9) = 3(9x^2 - 6x + 1)
or x^3 - 33 x^2 + 27x - 3 = 0
as it is cubic roots are tan^2 20,tan^2 40 tan^2 80
and sum of roots = - coefficient of x^2 or 33