Question

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

Answer 1

Given, $\tan (A+B)=\sqrt{3}$

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

$0^{\circ }<A+B\le {90}^{\circ }$ and $A>B$,

We know that, $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$ and $\tan {60}^{\circ }=\sqrt{3}$

Consider,

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

$⟹(A+B)={60}^{\circ }$ ---------------(i)

Consider,

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

$⟹(A-B)={30}^{\circ }$ ---------------(ii)

solve (i) and (ii),

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

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

By adding we get,

$2A={90}^{\circ }$

$A=\frac{{90}^{\circ }}{2}={45}^{\circ }$

We have, from (i) :

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

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

$B={60}^{\circ }-{45}^{\circ }$

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

$\therefore A={45}^{\circ },B={15}^{\circ }$