Question

If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}};0^{\circ }<A+B\le {90}^{\circ };A>B,$ find $A$ and $B.$

Answer 1

Given

$\tan (A+B)=\sqrt{3}⟶\left(1\right)$

$\tan (A-B)=\frac{1}{\sqrt{3}}⟶\left(2\right)$

and $A,Bϵ[0,{90}^0]$

and $A>B$

From $\left(1\right)$

$\tan (A+B)=\tan {60}^0$

$\Rightarrow A+B={60}^0⟶\left(3\right)$

From $\left(2\right)$

$\tan (A-B)=\frac{1}{\sqrt{3}}$

$\tan (A-B)=\tan {30}^0$

$\Rightarrow A-B={30}^0⟶\left(4\right)$

Solving $\left(3\right)$ & $\left(4\right)$ we get

$A+B={60}^0$

$\underset{–}{A-B={30}^0}$

$2A={90}^0$

$A={45}^0$

$B={15}^0$