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If $A$ and $B$ are acute angles satisfying $3 \cos ^2 A+2 \cos ^2 B=4$ and $\frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A}$, then $A+2 B=$
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Verified Answer
The correct answer is:
$90^{\circ}$
We have,
$$
\begin{array}{rlrl}
& 3 \cos ^2 A+2 \cos ^2 B=4 \\
\Rightarrow & 3\left(1-\sin ^2 A\right)+2\left(1-\sin ^2 B\right) & =4 \\
\Rightarrow & 5-3 \sin ^2 A-2 \sin ^2 B & =4 \\
\Rightarrow & 3 \sin ^2 A+2 \sin ^2 B & =1 \\
\Rightarrow & & 3 \sin ^2 A & =1-2 \sin ^2 B \\
\Rightarrow & & 3 \sin ^2 A & =\cos 2 B
\end{array}
$$
Again,
$$
\begin{aligned}
& \frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A} \\
& \Rightarrow \quad 3 \sin A \cos A=2 \sin B \cos B \\
& \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \\
& =\cos A \cos 2 B-\sin A \sin 2 B \\
& =\cos A\left(3 \sin ^2 A\right)-\sin A\left(\frac{3}{2} \sin 2 A\right) \\
& =3 \cos A \sin ^2 A-3 \sin ^2 A \cos A=0 \\
& \therefore A+2 B=90^{\circ} \\
&
\end{aligned}
$$
Now, $\cos (A+2 B)$
$$
\begin{array}{rlrl}
& 3 \cos ^2 A+2 \cos ^2 B=4 \\
\Rightarrow & 3\left(1-\sin ^2 A\right)+2\left(1-\sin ^2 B\right) & =4 \\
\Rightarrow & 5-3 \sin ^2 A-2 \sin ^2 B & =4 \\
\Rightarrow & 3 \sin ^2 A+2 \sin ^2 B & =1 \\
\Rightarrow & & 3 \sin ^2 A & =1-2 \sin ^2 B \\
\Rightarrow & & 3 \sin ^2 A & =\cos 2 B
\end{array}
$$
Again,
$$
\begin{aligned}
& \frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A} \\
& \Rightarrow \quad 3 \sin A \cos A=2 \sin B \cos B \\
& \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \\
& =\cos A \cos 2 B-\sin A \sin 2 B \\
& =\cos A\left(3 \sin ^2 A\right)-\sin A\left(\frac{3}{2} \sin 2 A\right) \\
& =3 \cos A \sin ^2 A-3 \sin ^2 A \cos A=0 \\
& \therefore A+2 B=90^{\circ} \\
&
\end{aligned}
$$
Now, $\cos (A+2 B)$
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