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?
 A: Your answer is correct. For all $n \in \mathbb{Z}$:
$$\tan(A-B) = 1 \implies A-B = \frac{\pi}{4}+\pi n$$
Recall that secant, like cosine, is an even function. (Which you've apparently referred to as well, after the format editing.)
$$\sec(-\theta) = \sec(\theta)$$
$$\sec(A+B) = \frac{2}{\sqrt 3} \implies \cos(A+B) = \frac{\sqrt 3}{2} \implies \pm(A+B) = \frac{\pi}{6}+2\pi n$$
Therefore, there are $2$ cases, neither of which yields the book's answer.
Case $1$: With $+(A+B)$, by subtracting the two equations to eliminate $A$, the following equation is obtained.
$$-2B = \frac{\pi}{4}+\pi n-\bigg(\frac{\pi}{6}+2\pi n\bigg)$$
$$-2B = \frac{\pi}{4}+\pi n-\frac{\pi}{6}-2\pi n$$
$$-2B = \frac{\pi}{12}-\pi n$$
$$B = \frac{\pi n}{2}-\frac{\pi}{12}$$
The minimum here is $B = \frac{11\pi}{24}$.
Case $2$: With $-(A+B)$, by adding the two equations to eliminate $A$, the following equation is obtained.
$$-2B = \frac{\pi}{4}+\pi n+\bigg(\frac{\pi}{6}+2\pi n\bigg)$$
$$-2B = \frac{\pi}{4}+\pi n+\frac{\pi}{6}+2\pi n$$
$$-2B = \frac{5\pi}{12}+3\pi n$$
$$B = -\frac{3\pi n}{2}-\frac{5\pi}{24}$$
The minimum here is $B = \frac{7\pi}{24}$.
Thus, the minimum positive value of $B$, as you correctly found, is $\frac{7\pi}{24}$.
A: 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 $\pm$.
A: You correctly have
$$
A=B+\frac{\pi}{4}+n\pi
$$
Therefore $A+B=2B+\pi/4+n\pi$. Hence
$$
\cos\left(2B+\frac{\pi}{4}+n\pi\right)=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}
$$
Therefore either
$$
2B+\frac{\pi}{4}+n\pi=\frac{\pi}{6}+2m\pi
$$
or
$$
2B+\frac{\pi}{4}+n\pi=-\frac{\pi}{6}+2m\pi
$$
In the first case
$$
2B=-\frac{\pi}{12}+(2m-n)\pi
$$
which is minimal positive for $2m-n=1$, giving
$$
B=\frac{11}{24}\pi
$$
In the second case,
$$
2B=-\frac{5}{12}\pi+(2m-n)\pi
$$
that gives the minimal positive solution
$$
B=\frac{7}{24}\pi
$$
You are right and the book is wrong.
