For alpha, beta, gamma neq 0. If sin ^{-1} al | vedclass

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(A) Let $\sin ^{-1} \alpha = A, \sin ^{-1} \beta = B, \sin ^{-1} \gamma = C$. Then $A+B+C = \pi$,which implies $\sin A = \alpha, \sin B = \beta, \sin

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$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) Let $\sin ^{-1} \alpha = A, \sin ^{-1} \beta = B, \sin ^{-1} \gamma = C$. Then $A+B+C = \pi$,which implies $\sin A = \alpha, \sin B = \beta, \sin C = \gamma$. The given equation is $(\alpha+\beta+\gamma)(\alpha+\beta-\gamma) = 3\alpha\beta$. This can be written as $(\alpha+\beta)^2 - \gamma^2 = 3\alpha\beta$. Expanding this,we get $\alpha^2 + \beta^2 + 2\alpha\beta - \gamma^2 = 3\alpha\beta$,which simplifies to $\alpha^2 + \beta^2 - \gamma^2 = \alpha\beta$. Dividing by $2\alpha\beta$,we get $\frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$. Using the law of cosines in a triangle with sides $a, b, c$ where $\sin A = \alpha, \sin B = \beta, \sin C = \gamma$,we have $\cos C = \frac{\alpha^2 + \beta^2 - \gamma^2}{2\alpha\beta} = \frac{1}{2}$. Since $\cos C = \frac{1}{2}$,we have $C = \frac{\pi}{3}$. Therefore,$\gamma = \sin C = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.

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 $\gamma$ is equal to

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