Question.
If $\tan (\mathrm{A}+\mathrm{B})=\sqrt{3}$ and $\tan (\mathrm{A}-\mathrm{B})=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$
$0^{\circ}<\mathrm{A}+\mathrm{B} \leq 90^{\circ} ; \mathrm{A}>\mathrm{B}$, find $\mathrm{A}$ and $\mathrm{B}$.
If $\tan (\mathrm{A}+\mathrm{B})=\sqrt{3}$ and $\tan (\mathrm{A}-\mathrm{B})=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$
$0^{\circ}<\mathrm{A}+\mathrm{B} \leq 90^{\circ} ; \mathrm{A}>\mathrm{B}$, find $\mathrm{A}$ and $\mathrm{B}$.
Solution:
$\tan (\mathrm{A}+\mathrm{B})=\sqrt{\mathbf{3}} \quad \Rightarrow \mathrm{A}+\mathrm{B}=60^{\circ}$ ...(1)
$\tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}} \quad \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}$ ...(2)
Adding (1) and (2),
$2 \mathrm{~A}=90^{\circ} \Rightarrow \mathrm{A}=45^{\circ}$
Then from $(1), 45^{\circ}+\mathrm{B}=60^{\circ} \Rightarrow \mathrm{B}=15^{\circ}$
$\tan (\mathrm{A}+\mathrm{B})=\sqrt{\mathbf{3}} \quad \Rightarrow \mathrm{A}+\mathrm{B}=60^{\circ}$ ...(1)
$\tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}} \quad \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}$ ...(2)
Adding (1) and (2),
$2 \mathrm{~A}=90^{\circ} \Rightarrow \mathrm{A}=45^{\circ}$
Then from $(1), 45^{\circ}+\mathrm{B}=60^{\circ} \Rightarrow \mathrm{B}=15^{\circ}$
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