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?

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 $\frac{7\pi}{24}$.
