Question

How do you solve arctan(a2)−arctana−0.22=0?

Answer 1

Start with the trigonometric identity

tan(α+β+γ)=tanα+tanβ+tanγ−tanαtanβtanγ1−tanαtanβ−tanαtanγ−tanβtanγ.

Let α, β, γ be respectively the arctangents of a, b, and c. Then

a+b+c−abc1−ab−ac−bc=tanπ=0.

A fraction is 0 only if the numerator is 0. So we need

a+b+c=abc.

Since 1+2+3=1⋅2⋅3, we have at least one solution. I'm not actually sure if there are others.

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