JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 2)
For $$\alpha, \beta, \gamma \neq 0$$, if $$\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$$ and $$(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$$, then $$\gamma$$ equals
$$\sqrt{3}$$
$$\frac{\sqrt{3}}{2}$$
$$\frac{1}{\sqrt{2}}$$
$$\frac{\sqrt{3}-1}{2 \sqrt{2}}$$
Explanation
Let $$\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C$$
$$\begin{aligned} & \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ & (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\ & \alpha^2+\beta^2-\gamma^2=\alpha \beta \\ & \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\ & \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\ & \sin \mathrm{C}=\gamma \\ & \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\ & \gamma=\frac{\sqrt{3}}{2} \end{aligned}$$
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