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If $\triangle A B C$, if $\cos A \cos B+\sin A \sin B \sin C=1$ and $C=\frac{\pi}{2}$, then $A: B=$
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Verified Answer
The correct answer is:
$1: 1$
Given,
$$
\begin{aligned}
& \cos A \cos B+\sin A \sin B \sin C=1 \text { and } C=\frac{\pi}{2} \\
& \Rightarrow \quad \sin C=\sin \frac{\pi}{2}=1
\end{aligned}
$$
So, $\cos A \cos B+\sin A \sin B(\mathrm{l})=\mathrm{l}$
$$
\begin{array}{cc}
\Rightarrow & \cos (A-B)=1 \\
\Rightarrow & {[\because \cos (A+B)=\cos A \cos B-\sin A \sin B]} \\
\Rightarrow & 1-\cos (A-B)=0 \\
& 2 \sin ^2\left(\frac{A-B}{2}\right)=0
\end{array}
$$
It is only possible, when
$$
\begin{array}{rlrl}
& & A-B & =0 \\
\Rightarrow & A & =B \\
\Rightarrow & \frac{A}{B} & =1
\end{array}
$$
Hence, $A: B=1: 1$
$$
\begin{aligned}
& \cos A \cos B+\sin A \sin B \sin C=1 \text { and } C=\frac{\pi}{2} \\
& \Rightarrow \quad \sin C=\sin \frac{\pi}{2}=1
\end{aligned}
$$
So, $\cos A \cos B+\sin A \sin B(\mathrm{l})=\mathrm{l}$
$$
\begin{array}{cc}
\Rightarrow & \cos (A-B)=1 \\
\Rightarrow & {[\because \cos (A+B)=\cos A \cos B-\sin A \sin B]} \\
\Rightarrow & 1-\cos (A-B)=0 \\
& 2 \sin ^2\left(\frac{A-B}{2}\right)=0
\end{array}
$$
It is only possible, when
$$
\begin{array}{rlrl}
& & A-B & =0 \\
\Rightarrow & A & =B \\
\Rightarrow & \frac{A}{B} & =1
\end{array}
$$
Hence, $A: B=1: 1$
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