Question

If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}};0^o<A+B\le {90}^o;A>B$ then the value of A and B

Answer 1

Given:-

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

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

$0°<A+B\le 90°;A>B$

To find:- $A=?B=?$

Solution:-

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

$\Rightarrow \tan (A+B)=\tan 60°$

$\Rightarrow A+B=60°.....\left(1\right)$

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

$\Rightarrow \tan (A-B)=\tan 30°$

$\Rightarrow A-B=30°.....\left(2\right)$

Adding equatio $\left(1\right)\&\left(2\right)$, we have

$(A+B)+(A-B)=60°+30°$

$\Rightarrow 2A=90°$

$\Rightarrow A=45°$

Substituting the value of $A$ in equation $\left(1\right)$, we have

$45°+B=60°$

$\Rightarrow B=60°-45°=15°$

Hence the value of $A$ and $B$ is $45v$ and $15°$ respectively.