<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] 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.
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