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If $\cot \alpha \cot \beta=2$, then $\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{3}$
We have,
$\cot \alpha \cot \beta=2$
$$
\begin{aligned}
&\text { Now, } \frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\cos \alpha \cos \beta+\sin \alpha \sin \beta} \\
&=\frac{\cot \alpha \cot \beta-1}{\cot \alpha \cot \beta+1}=\frac{2-1}{2+1}=\frac{1}{3}
\end{aligned}
$$
$\cot \alpha \cot \beta=2$
$$
\begin{aligned}
&\text { Now, } \frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\cos \alpha \cos \beta+\sin \alpha \sin \beta} \\
&=\frac{\cot \alpha \cot \beta-1}{\cot \alpha \cot \beta+1}=\frac{2-1}{2+1}=\frac{1}{3}
\end{aligned}
$$
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