Question

If $\tan \left(\mathrm{A}+\mathrm{B}\right)=\sqrt{3},\tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{3}}$ and $0⩽\mathrm{A}+\mathrm{B}\le \frac{\mathrm{\pi }}{2},\mathrm{A}>\mathrm{B}$ . Find $\mathrm{A}$ and $\mathrm{B}$.

Answer 1

Step 1: Form linear equations from the given equations

Given : $\tan \left(\mathrm{A}+\mathrm{B}\right)=\sqrt{3},\tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{3}}$ and $0⩽\mathrm{A}+\mathrm{B}\le \frac{\mathrm{\pi }}{2},\mathrm{A}>\mathrm{B}$

Since $0⩽\mathrm{A}+\mathrm{B}\le \frac{\mathrm{\pi }}{2}$ both $\mathrm{A}$ and $\mathrm{B}$ lie in first quadrant.

$\begin{array}{ccc}& \tan \left(\mathrm{A}+\mathrm{B}\right)=\sqrt{3}& \\ \Rightarrow & \mathrm{A}+\mathrm{B}=60°& ...\left(\mathrm{i}\right)\\ & \tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{1}{\sqrt{3}}& \\ \Rightarrow & \mathrm{A}-\mathrm{B}=30°& ...\left(\mathrm{ii}\right)\end{array}$

Step 2: Simplify the linear equations

Add $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$.

$\begin{array}{cc}& \mathrm{A}+\mathrm{B}+\mathrm{A}-\mathrm{B}=60+30\\ \Rightarrow & 2\mathrm{A}=90\\ \Rightarrow & \mathrm{A}=45°\end{array}$

Substitute in $\left(\mathrm{i}\right)$

$\begin{array}{cc}& \mathrm{A}+\mathrm{B}=60\\ \Rightarrow & 45+\mathrm{B}=60\\ \Rightarrow & \mathrm{B}=15°\end{array}$

Hence, $\mathrm{A}=45°$ and $\mathrm{B}=15°$.