Htykuut

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 $A-B=npi+pi/4$ and $A+B = 2npipmpi/6$. Solving these I am getting $B =7pi/24$ and $A =37pi/24$.

The book I am refering to has marked the answer as $19pi/24$.

Am I doing something wrong?

You correctly have
$$
A=B+frac{pi}{4}+npi
$$
Therefore $A+B=2B+pi/4+npi$. Hence
$$
cosleft(2B+frac{pi}{4}+npiright)=frac{sqrt{3}}{2}=cosfrac{pi}{6}
$$
Therefore either
$$
2B+frac{pi}{4}+npi=frac{pi}{6}+2mpi
$$
or
$$
2B+frac{pi}{4}+npi=-frac{pi}{6}+2mpi
$$
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.

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 $secleft(thetaright) = frac{2}{sqrt{3}}$; 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 $frac{7pi}{24}$.

Explicitly, $$A-B = frac{pi}{4} + npi,$$ $$A+B = pm frac{pi}{6} + 2mpi,$$ so $$B = frac{pi}{2}left(pmfrac{1}{6} - frac{1}{4} + left(2m - nright)right),$$ and then we see that $$B = frac{pi}{2}left(-frac{1}{12} + kright), quad k in mathbb{Z}$$ or $$B = frac{pi}{2}left(frac{-5}{12} + kright), quad k in mathbb{Z},$$ and the smallest positive value of $B$ is then $frac{7pi}{24}$.

EDIT: I see that you have/somebody has now edited your question to include the $pm$.

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.)

$$sec(-theta) = sec(theta)$$

$$sec(A+B) = frac{2}{sqrt 3} implies cos(A+B) = frac{sqrt 3}{2} implies pm(A+B) = frac{pi}{6}+2pi n$$

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.

$$-2B = frac{pi}{4}+pi n-bigg(frac{pi}{6}+2pi nbigg)$$

$$-2B = frac{pi}{4}+pi n-frac{pi}{6}-2pi n$$

$$-2B = frac{pi}{12}-pi n$$

$$B = frac{pi n}{2}-frac{pi}{12}$$

The minimum here is $B = frac{11pi}{24}$.

Case $2$: With $-(A+B)$, by adding the two equations to eliminate $A$, the following equation is obtained.

$$-2B = frac{pi}{4}+pi n+bigg(frac{pi}{6}+2pi nbigg)$$

$$-2B = frac{pi}{4}+pi n+frac{pi}{6}+2pi n$$

$$-2B = frac{5pi}{12}+3pi n$$

$$B = -frac{3pi n}{2}-frac{5pi}{24}$$

The minimum here is $B = frac{7pi}{24}$.

Thus, the minimum positive value of $B$, as you correctly found, is $frac{7pi}{24}$.