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If $tan\ (A + B) = \sqrt3$ and $tan\ (A - B) = \frac{1}{\sqrt3}$; $0^o < A + B ≤ 90^o; A > B$, find $A$ and $B$.
Given:
$tan\ (A + B) = \sqrt3$ and $tan\ (A - B) = \frac{1}{\sqrt3}$; $0^o < A + B ≤ 90^o; A > B$
To do:
We have to find $A$ and $B$.
Solution:  
$\tan (A-B)=\frac{1}{\sqrt3}$
$\tan (A-B)=\tan 30^{\circ}$ (Since $\tan 30^{\circ}=\frac{1}{\sqrt3}$)
$\Rightarrow A-B=30^{\circ}$......(i)
$\tan (A+B)=\sqrt3$
$\tan (A+B)=\tan 60^{\circ}$ (Since $\tan 60^{\circ}=\sqrt3$)
$\Rightarrow A+B=60^{\circ}$
$\Rightarrow A=60^{\circ}-B$........(ii)
Substituting (ii) in (i), we get,
$60^{\circ}-B-B=30^{\circ}$
$\Rightarrow 2B=30^{\circ}$
$\Rightarrow B=\frac{30^{\circ}}{2}$
$\Rightarrow B=15^{\circ}$
$\Rightarrow A=60^{\circ}-15^{\circ}=45^{\circ}$
The values of $A$ and $B$ are $45^{\circ}$ and $15^{\circ}$ respectively.  
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