The equation is tan(30+θ) = 2tan(60-θ), how can it be written in the form of tan^2θ + (6√ 3)tanθ - 5 = 0?

See the derivation in the Explanation Section below.

Suppose that, ##(30+theta)=alpha, &, (60-theta)=beta##, so that,

##alpha+beta=90, or, beta=90-alpha##.

Sub.ing these in the given eqn., we have,

##tanalpha=2tanbeta=2tan(90-alpha)=2cotalpha=2/tanalpha##, or,

##tan^2alpha=2##

##:. tan^2(30+theta)=2##

##:. ((tan30+tantheta)/(1-tan30*tantheta))^2=2##.

##:. ((1/sqrt3+t)/(1-1/sqrt3*t))^2=2, "where, "t=tantheta##

##:. (1+sqrt3*t)^2=2(sqrt3-t)^2##.

##:. 1+2sqrt3*t+3t^2=6-4sqrt3*t+2t^2##.

##:. t^2+6sqrt3*t-5=0##, i.e.,

##tan^2theta+6sqrt3tantheta-5=0##, as desired!

Enjoy Maths.!.