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If $\cot \mathrm{A}=2$ and $\cot \mathrm{B}=3$, then what is the value of $\mathrm{A}+\mathrm{B}$ ?
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Verified Answer
The correct answer is:
$\pi / 4$
\cot \mathrm{A}=2 \text { and } \cot \mathrm{B}=3\\
\begin{aligned}
& \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}=\frac{6-1}{2+3}=\frac{5}{5}=1 \\
& \Rightarrow \cot (A+B)=\cot \left(\frac{\pi}{4}\right) \Rightarrow A+B=\frac{\pi}{4} \\
&=\left[\sin \left(90^{\circ}-23 \frac{1^{\circ}}{2}\right)\right]^{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos ^{2} 23 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos 2\left(23 \frac{1^{\circ}}{2}\right)=\cos 47^{\circ} \\
\left(\because \cos 2 \mathrm{~A}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}\right) \\
\begin{aligned}
& \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}=\frac{6-1}{2+3}=\frac{5}{5}=1 \\
& \Rightarrow \cot (A+B)=\cot \left(\frac{\pi}{4}\right) \Rightarrow A+B=\frac{\pi}{4} \\
&=\left[\sin \left(90^{\circ}-23 \frac{1^{\circ}}{2}\right)\right]^{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos ^{2} 23 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos 2\left(23 \frac{1^{\circ}}{2}\right)=\cos 47^{\circ} \\
\left(\because \cos 2 \mathrm{~A}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}\right) \\
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