Question

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

Answer 1

$\tan (A+B)=\sqrt{3}=\tan {60}^{\circ }$

$A+B={60}^{\circ }$.......I

$\tan (A-B)=\frac{1}{\sqrt{3}}=\tan {30}^{\circ }$

$A-B={30}^{\circ }$.........II

solving I and II, we get,

$A={45}^{\circ }$

$B={15}^{\circ }$