For α,β,γ≠0. If sin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ, then γ equal to
Question
For α,β,γ≠0. If sin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ, then γ equal to
Solution
The given conditions are sin^(-1)α + sin^(-1)β + sin^(-1)γ = π and (α+β+γ)(α−γ+β)=3αβ.
From the first condition, we can infer that α, β, and γ are the sides of a triangle (since the sum of the angles in a triangle is π). Therefore, we have α^2 + β^2 + γ^2 = 1 (since the square of the sides of a triangle is equal to 1).
From the second condition, we can expand it to get α^2 + β^2 + γ^2 + 2αβ - 2γα + 2βγ = 3αβ.
Substituting α^2 + β^2 + γ^2 = 1 into the equation, we get 1 + 2αβ - 2γα + 2βγ = 3αβ.
Solving for γ, we get γ = (1 - αβ) / (α - β).
Therefore, the value of γ is (1 - αβ) / (α - β).
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