tanθ+β =pi/4, show that (1+tanθ)(1+tan )=2

Let

tanֿ¹(1/2) = α tanֿ¹(1/3) = β α + β = θ tanθ = tan(α + β) = [tanα + tanβ] / [1 - tanαtanβ] = [(1/2) + (1/3)] / [1 - (1/2)(1/3)] = [3 + 2] / [6 - 1] = 1 Angles α and β are both acute, so α + β < π, θ < π. This means θ must be in quadrant 1 or 2. tanθ = 1 θ = π/4 α + β = π/4 tanֿ¹(1/2) + tanֿ¹(1/3) = π/4

so from above we know that

tanθ = 1 by putting

(1+tanθ)(1+tan )

(1+1)(1+0)=2

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