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 [JEE Main 2024, 31 Jan (Shift 1)]

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Correct answer: b) $\frac{\sqrt{3}}{2}$. $A=\sin^{-1}\alpha,\quad B=\sin^{-1}\beta,\quad C=\sin^{-1}\gamma$ $A+B+C=\pi$ $\gamma=\sin C=\sin(A+B)$ $=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$ $(\alpha+\beta+\gamma)(\alpha+\beta-\gamma)=3\alpha\beta$ $(\alpha+\beta)^2-\gamma^2=3\alpha\beta$ $\gamma=\sin(A+B)$ $(\sin A+\sin B)^2-\sin^2(A+B)=3\sin A\sin B$ $\sin(A+B)=\sin A\cos B+\cos A\sin B$ $\sin A\sin B\bigl(1+\cos(A+B)\bigr)$ $\sin A\sin B\bigl(1+\cos(A+B)\bigr)=3\sin A\sin B$ Since $\alpha,\beta\ne0$ $\sin A\sin B\ne0$ $1+\cos(A+B)=3$ $\cos(A+B)=\dfrac12$ $A+B=\pi-C$ $\cos(A+B)=\cos(\pi-C)=-\cos C$ $-\cos C=\dfrac12$ $\cos C=-\dfrac12$ $C\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ $(\alpha+\beta)^2-\gamma^2=3\alpha\beta$ $\alpha^2+\beta^2-\alpha\beta=\gamma^2$ $\gamma=\sin(A+B)$ $\sin^2(A+B)=\alpha^2+\beta^2-\alpha\beta$ $\sin^2(A+B)=(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})^2$ $2\sqrt{(1-\alpha^2)(1-\beta^2)}=1$ $(1-\alpha^2)(1-\beta^2)=\dfrac14$ $\cos(A+B)=\sqrt{1-\alpha^2}\sqrt{1-\beta^2}-\alpha\beta$ \(=\df

For \(\alpha, \beta, \gamma \neq 0\), if \(\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi\) and \((\alpha+\bet…

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