$If \tan \left(A-B\right)=1,\sec \left(A+B\right)=\frac{2}{\sqrt{3}}, then least positive values of A, B are$

$$\begin{gathered} We have \tan \left(A-B\right)=1\Rightarrow \tan \left(A-B\right)=\tan \frac{{\displaystyle \pi }}{{\displaystyle 4}} \\ \Rightarrow A-B=n\pi +\frac{\pi }{4} \end{gathered}$$

$\Rightarrow A-B=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{9\mathrm{\pi }}{4},....$

$$\begin{gathered} Also \sec \left(A+B\right)=\frac{2}{\sqrt{3}}\Rightarrow \cos \left(A+B\right)=\frac{\sqrt{3}}{2}=\cos \frac{\mathrm{\pi }}{6} \\ \Rightarrow A+B=2n\mathrm{\pi }\pm \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 6}}=\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 6}},\frac{{\displaystyle 11\mathrm{\pi }}}{{\displaystyle 6}},\frac{{\displaystyle 13\mathrm{\pi }}}{{\displaystyle 6}}.... \end{gathered}$$

$$\begin{gathered} Since A and B are to be the least positive values, we must have \\ A-B=\frac{\mathrm{\pi }}{4} and A+B=\frac{11\mathrm{\pi }}{6} \\ \Rightarrow 2A=\frac{50\mathrm{\pi }}{24}\Rightarrow A=\frac{25\mathrm{\pi }}{24} \\ Also 2B=\frac{38\mathrm{\pi }}{24}\Rightarrow B=\frac{19\mathrm{\pi }}{24} \end{gathered}$$