Verified Solution • 2025–26 Session moderate QJEEM24Madd54: JEE Main 2024 Mathematics - Class 12 JEEM24Madd54 Solution: For , if and , then equals For α,β,γ≠0\alpha, \beta, \gamma \neq 0α,β,γ=0 , if sin−1α+sin−1β+sin−1γ=π\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pisin−1α+sin−1β+sin−1γ=π and (α+β+γ)(α−γ+β)=3αβ(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta(α+β+γ)(α−γ+β)=3αβ , then γ\gammaγ equals Multiple Choice Options A 3\sqrt{3}3 B 32\frac{\sqrt{3}}{2}23 C 12\frac{1}{\sqrt{2}}21 D 3−122\frac{\sqrt{3}-1}{2 \sqrt{2}}223−1 Step-by-Step Verified Solution Let sin−1α=A,sin−1β=B,sin−1γ=C\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=Csin−1α=A,sin−1β=B,sin−1γ=C amp;A+B+C=πamp;(α+β)2−γ2=3αβamp;α2+β2−γ2=αβamp;α2+β2−γ22αβ=12amp;⇒cosC=12amp;sinC=γamp;cosC=1−γ2=12amp;γ=32\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}amp;A+B+C=πamp;(α+β)2−γ2=3αβamp;α2+β2−γ2=αβamp;2αβα2+β2−γ2=21amp;⇒cosC=21amp;sinC=γamp;cosC=1−γ2=21amp;γ=23 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring