<s>[INST] If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$ 
If $\tan(A-B)=1$ and $\sec(A+B)=2/\sqrt{3}$, find the minimum positive value of $B$

I am using $\tan(A-B)=1$, so $A-B=n\pi+\pi/4$ and $A+B = 2n\pi\pm\pi/6$. Solving these I am getting $B =7\pi/24$ and $A =37\pi/24$.
The book I am refering to has marked the answer as $19\pi/24$.
Am I doing something wrong?
 [/INST] First of all, your values for $A$ and $B$ satisfy the given formulae and $B$ is positive, so their proposed minimum for $B$ is definitely wrong.
One thing you did miss in your reasoning is that there are two families of solutions to $\sec\left(\theta\right) = \frac{2}{\sqrt{3}}$; there should be a $\pm$ with $\frac{\pi}{6}$ in your formula for $A+B$. However, it turns out that you get the minimum value for $B$ from the family of solutions you gave, and the minimum positive value of $B$ is indeed $\frac{7\pi}{24}$.
Explicitly, $$A-B = \frac{\pi}{4} + n\pi,$$ $$A+B = \pm \frac{\pi}{6} + 2m\pi,$$ so $$B = \frac{\pi}{2}\left(\pm\frac{1}{6} - \frac{1}{4} + \left(2m - n\right)\right),$$ and then we see that $$B = \frac{\pi}{2}\left(-\frac{1}{12} + k\right), \quad k \in \mathbb{Z}$$ or $$B = \frac{\pi}{2}\left(\frac{-5}{12} + k\right), \quad k \in \mathbb{Z},$$ and the smallest positive value of $B$ is then $\frac{7\pi}{24}$.
EDIT: I see that you have/somebody has now edited your question to include the $\pm$.
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