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If $\cos (A + B) = \frac{1}{2}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$, where $0 \leq A + B \leq 90^\circ$, then find the value of $\sec (2A-3B)$.
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$\cos(A + B) = \frac{1}{2} \Rightarrow A + B = 60^\circ$ ... (i)
$\tan(A - B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30^\circ$ ... (ii)
Solving (i) and (ii), we get $A = 45^\circ$ and $B = 15^\circ$
$\Rightarrow \sec(2A - 3B) = \sec(90^\circ - 45^\circ)$
$= \sec 45^\circ = \sqrt{2}$
$\tan(A - B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30^\circ$ ... (ii)
Solving (i) and (ii), we get $A = 45^\circ$ and $B = 15^\circ$
$\Rightarrow \sec(2A - 3B) = \sec(90^\circ - 45^\circ)$
$= \sec 45^\circ = \sqrt{2}$