Help me plz

Help me plz Q No.; 61 3Sin A If A and B are satisfying the 3 • 2 g • 4 and sin 2:osB . then A. 2B

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Please find the below required solution.

$$\begin{gathered} \mathrm{We} \mathrm{are} \mathrm{given} \\ 3 {\cos }^2\mathrm{A}+2{\cos }^2\mathrm{B}=4 ........\left(\mathrm{i}\right) \\ \mathrm{and} \\ \frac{3\mathrm{sinA}}{\mathrm{sinB}}=\frac{2\mathrm{cosB}}{\mathrm{cosA}} \\ \Rightarrow 3 \mathrm{sinA} \mathrm{cosA}=2 \mathrm{sinB} \mathrm{cosB} \\ \Rightarrow 3 \sin 2\mathrm{A}=2 \sin 2\mathrm{B} ............\left(\mathrm{ii}\right) \\ \mathrm{Now}, \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \cos 2\mathrm{B}-\mathrm{sinA} \sin 2\mathrm{B} \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \cos 2\mathrm{B}-\mathrm{sinA} \left(\frac{3}{2}\sin 2\mathrm{A}\right) \left[\mathrm{From} \mathrm{equation} \left(\mathrm{ii}\right)\right] \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \cos 2\mathrm{B}-\mathrm{sinA}\left(3 \mathrm{sinA} \mathrm{cosA}\right) \left[\mathrm{As}, \sin 2\mathrm{A}=2 \mathrm{sinA} \mathrm{cosA}\right] \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \left\{\cos 2\mathrm{B}-3 {\sin }^2\mathrm{A}\right\} \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \left\{2 {\cos }^2\mathrm{B}-1-3\left(1-{\cos }^2\mathrm{A}\right)\right\} \left[\mathrm{As}, \cos 2\mathrm{x}=2 {\cos }^2\mathrm{x}-1 \mathrm{and} {\sin }^2\mathrm{x}=1-{\cos }^2\mathrm{x}\right] \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA} \left\{\left(4-3{\cos }^2\mathrm{A}\right)-1-3\left(1-{\cos }^2\mathrm{A}\right)\right\} \left[\mathrm{From} \mathrm{equation} \left(\mathrm{i}\right), 2{\cos }^2\mathrm{B}=4-3{\cos }^2\mathrm{A}\right] \\ \Rightarrow \cos \left(\mathrm{A}+2\mathrm{B}\right)=\mathrm{cosA}\left\{4-3{\cos }^2\mathrm{A}-1-3+3{\cos }^2\mathrm{A}\right\}=\mathrm{cosA}\times \left(0\right)=0 \\ \mathrm{As}, \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} \mathrm{positive} \mathrm{acute} \mathrm{angles}, \mathrm{therefore} 0<\mathrm{A}+2\mathrm{B}<\mathrm{\pi }. \\ \mathrm{Hence}, \cos \left(\mathrm{A}+2\mathrm{B}\right)=0 \Rightarrow \mathrm{A}+2\mathrm{B}=\frac{\mathrm{\pi }}{2}. \\ \mathrm{Therefore}, \mathrm{option} \left(\mathrm{b}\right) \mathrm{is} \mathrm{the} \mathrm{correct} \mathrm{option}. \end{gathered}$$

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