Question

$\tan (a-b)=\frac{1}{\sqrt{3}}$ and $\tan (a+b)=\sqrt{3}$ find a and b.

Answer 1

Given that,

$\tan (a-b)=\frac{1}{\sqrt{3}}$ and $\tan (a+b)=\sqrt{3}$

$\because \tan {30}^{\circ }=\frac{1}{\sqrt{3}}$ and $\tan {60}^{\circ }=\sqrt{3}$

So, $a-b={30}^{\circ }$ and $a+b={60}^{\circ }$

By sovling we get,

$a={45}^{\circ }$ and $b={15}^{\circ }$