This question was uploaded on 07/01/22 on social media accounts.
This question is from IIT JEE Advanced examination.
Solution:
From the first equation:
\[\color{red} {3\sin^2A+2\sin^2B=1}\]
\[\Rightarrow\color{red} {3\sin^2A=1-2\sin^2B=\cos2B}\]
\[\Rightarrow\color{red} {3\sin A=\frac{\cos2B}{\sin A}}\]
From the second equation:
\[\color{blue} {3\sin2A=2\sin2B = 3(2\sin A \cos A)}\]
\[\Rightarrow\color{blue} {3\sin A=\frac{\sin2B}{\cos A}}\]
From both equations:
\[\color{red} {\frac{\cos2B}{\sin A}}=\color{blue} {\frac{\sin2B}{\cos A}}\]
\[\Rightarrow\color{blue}{\cos A}\color{red}{\cos2B}-\color{red}{\cos A}\color{blue}{\sin2B}=0\]
\[\Rightarrow\cos(A+2B)=0\]
\[\Rightarrow A+2B=\frac\pi2\]
