Question:
If $\tan (A-B)=1$ and $\sec (A+B)=\frac{2}{\sqrt{3}}$, the smallest positive value of $B$ is
(a) $\frac{25 \pi}{24}$
(b) $\frac{19}{24}$
(c) $\frac{13 \pi}{24}$
(d) $\frac{11 \pi}{24}$
Solution:
(b) $\frac{19}{24}$
Given:
$\tan (A-B)=1$ and $\sec (A+B)=\frac{2}{\sqrt{3}}$
$\Rightarrow A-B=\frac{\pi}{4} \ldots(1)$ and $A+B=\frac{\pi}{6} \ldots(2)$
Adding these equations we get:
$2 A=\frac{\pi}{4}+\frac{\pi}{6}$
$\Rightarrow A=\frac{5 \pi}{24}$
$\Rightarrow$ Smallest possible value of $B=\pi-\frac{5 \pi}{24}=\frac{19 \pi}{24}$.
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