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Mathematics Question on Trigonometry

For α,β,γ≠0\alpha, \beta, \gamma \neq 0. If sin⁡−1α+sin⁡−1β+sin⁡−1γ=π\sin^{-1} \alpha + \sin^{-1} \beta + \sin^{-1} \gamma = \pi and (α+β+γ)(α−γ+β)=3αβ(\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha \beta, then γ\gamma is equal to

A

32\frac{\sqrt{3}}{2}

B

12\frac{1}{\sqrt{2}}

C

3−122\frac{\sqrt{3} - 1}{2\sqrt{2}}

D

3\sqrt{3}

Answer

32\frac{\sqrt{3}}{2}

Explanation

Solution

Let sin⁡−1α=A\sin^{-1} \alpha = A, sin⁡−1β=B\sin^{-1} \beta = B, sin⁡−1γ=C\sin^{-1} \gamma = C

A+B+C=πA + B + C = \pi

(α+β)2−γ2=3αβ(\alpha + \beta)^2 - \gamma^2 = 3 \alpha \beta

α2+β2−γ2=αβ\alpha^2 + \beta^2 - \gamma^2 = \alpha \beta

α2+β2−γ22αβ=12\frac{\alpha^2 + \beta^2 - \gamma^2}{2 \alpha \beta} = \frac{1}{2}

⇒cos⁡C=12\Rightarrow \cos C = \frac{1}{2}

sin⁡C=γ\sin C = \gamma

cos⁡C=1−γ2=12\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}

γ=32\gamma = \frac{\sqrt{3}}{2}