If tan(A − B) = 1 and sec(A + B) = 2/√3, Find the Smallest Positive Value of B

Solution

Given,

\[ \tan(A-B)=1 \]

Therefore,

\[ A-B=\frac{\pi}{4} \]

Also,

\[ \sec(A+B)=\frac{2}{\sqrt3} \]

Hence,

\[ \cos(A+B)=\frac{\sqrt3}{2} \]

Therefore,

\[ A+B=\frac{\pi}{6} \]

or

\[ A+B=\frac{11\pi}{6} \]

Taking the value that gives positive \(B\),

\[ A+B=\frac{11\pi}{6} \]

Now subtract:

\[ (A+B)-(A-B) = \frac{11\pi}{6}-\frac{\pi}{4} \]

\[ 2B = \frac{22\pi-3\pi}{12} \]

\[ 2B = \frac{19\pi}{12} \]

Therefore,

\[ B=\frac{19\pi}{24} \]

Final Answer

\[ \boxed{ B=\frac{19\pi}{24} } \]

Correct Option: (b)