For α, β, γ ≠ 0, if sin ⁻¹ α+sin ⁻¹ β+sin ⁻¹ γ=π and (α+β+γ)(α-γ+β)=3 α β, then γ equals

Solution

Maths sir · 4 steps

Three inverse sines adding to π means three angles of a triangle, so read the condition as a cosine rule.

1. Angles of a triangle. Let A = sin⁻¹ α, B = sin⁻¹ β, C = sin⁻¹ γ, each in [-π/2, π/2]. Any two add to at most π, so A + B + C = π forces each angle to be ≥ 0. They are nonzero, so A, B, C are the angles of a triangle, with sin A = α, sin B = β, sin C = γ.
2. Sides follow sines. By the sine rule the sides are a : b : c = α : β : γ. The given condition has the same degree in every term, so you may write it with a, b, c in place of α, β, γ.
3. Expand the condition. (α + β + γ)(α + β - γ) = (α + β)² - γ². Setting this equal to 3αβ gives α² + β² - γ² = αβ.

   cos C = (a² + b² - c²)/2ab = (α² + β² - γ²)/2αβ = αβ/2αβ = 1/2

4. So. C = π/3, and γ = sin C = √3/2, which is option (b).

The idea to keep. When sin⁻¹ or cos⁻¹ terms add to π, think of a triangle. Sines become sides by the sine rule, and a condition like a² + b² - c² = kab is a cosine rule in disguise.

Answer from the NTA final answer key. Solution by PrepWiser.