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If cotA $\cot B=2$, then what is the value of $\cos (\mathrm{A}+\mathrm{B}) \sec (\mathrm{A}-\mathrm{B}) ?$
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Verified Answer
The correct answer is:
$\frac{1}{3}$
Consider $\cos (\mathrm{A}+\mathrm{B}) \cdot \sec (\mathrm{A}-\mathrm{B})$
$=\frac{\cos (\mathrm{A}+\mathrm{B})}{\cos (\mathrm{A}-\mathrm{B})}=\frac{\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}}{\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}}$
Divide $\mathrm{Nr}$ and $\mathrm{Dr}$ by $\sin \mathrm{A} \sin \mathrm{B}$,
$\frac{\cot \mathrm{A} \cot \mathrm{B}-1}{\cot \mathrm{A} \cot \mathrm{B}+1}=\frac{2-1}{2+1}=\frac{1}{3}$
$=\frac{\cos (\mathrm{A}+\mathrm{B})}{\cos (\mathrm{A}-\mathrm{B})}=\frac{\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}}{\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}}$
Divide $\mathrm{Nr}$ and $\mathrm{Dr}$ by $\sin \mathrm{A} \sin \mathrm{B}$,
$\frac{\cot \mathrm{A} \cot \mathrm{B}-1}{\cot \mathrm{A} \cot \mathrm{B}+1}=\frac{2-1}{2+1}=\frac{1}{3}$
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