Question

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

• A. $A={35}^{\circ }B={20}^{\circ }$
• B. $A={60}^{\circ }B={20}^{\circ }$
• C. $A={30}^{\circ }B={75}^{\circ }$
• D. $A={45}^{\circ }B={15}^{\circ }$

Answer 1

The correct option is D $A={45}^{\circ }B={15}^{\circ }$

$\tan (A+B)=\sqrt{3}\Rightarrow A+B={60}^{\circ }...\left(1\right)$
$\tan (A-B)=\frac{1}{\sqrt{3}}\Rightarrow A+B={30}^{\circ }...\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$
$2A={90}^{\circ }\Rightarrow A={45}^{\circ }$
Then, from $\left(1\right)$, ${45}^{\circ }+B={60}^{\circ }\Rightarrow B={15}^{\circ }$