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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.!.

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