Thank you for reporting, we will resolve it shortly
Q.
If $tan(A - B) = 1$, $sec\left(A+B\right)=\frac{2}{\sqrt{3}}$, the smallest positive value of $B$ is
Trigonometric Functions
Solution:
$tan(A - B) = 1$
$\Rightarrow A-B=45^{\circ}$ or $225^{\circ}$
$sec\left(A+B\right)=\frac{2}{\sqrt{3}}$
$\Rightarrow A + B = 30^{\circ}$ or $330^{\circ}$
$A+B=330^{\circ}=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\quad\ldots\left(i\right)$
and $A-B=225^{\circ}=\frac{5\pi}{4}\quad\ldots\left(ii\right)$
Solving $(i)$ and $(ii)$, we get
$2B=\frac{11\pi}{6}-\frac{5\pi}{4}$
$\Rightarrow 2B=\frac{7\pi}{12}$
$\Rightarrow B=\frac{7\pi}{24}$