Question

If tan (A − B) = 1 and sec (A + B) = $\frac{2}{\sqrt{3}}$, the smallest positive value of B is

(a) $\frac{25\mathrm{\pi }}{24}$
(b) $\frac{19\mathrm{\pi }}{24}$
(c) $\frac{13\mathrm{\pi }}{24}$
(d) $\frac{11\mathrm{\pi }}{24}$

Answer 1

(b) $\frac{19\mathrm{\pi }}{24}$
$$\begin{gathered} \text{Given}: \\ \tan (A-B)=1\mathrm{and}\sec (A+B)=\frac{2}{\sqrt{3}} \\ \Rightarrow A-B=\frac{\mathrm{\pi }}{4}...\left(1\right)\mathrm{and}A+B=\frac{\mathrm{\pi }}{6}...\left(2\right) \\ \mathrm{Adding}\mathrm{these}\mathrm{equations}\mathrm{we}\mathrm{get}: \\ 2A=\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6} \\ \Rightarrow A=\frac{5\mathrm{\pi }}{24} \\ \Rightarrow \mathrm{Smallest}\mathrm{possible}\mathrm{value}\mathrm{of}B=\mathrm{\pi }-\frac{5\mathrm{\pi }}{24}=\frac{19\mathrm{\pi }}{24}. \end{gathered}$$