$I f \tan (A - B) = 1, \sec (A + B) = \frac{2}{\sqrt{3}}, t h e n l e a s t p o s i t i v e v a l u e s o f A, B a r e$

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$\frac{25 π}{24}, \frac{19 π}{24}$

$\frac{13 π}{24}, \frac{31 π}{24}$

$\frac{19 π}{24}, \frac{25 π}{24}$

$\frac{31 π}{24}, \frac{13 π}{24}$

answer is A.

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Detailed Solution

$W e h a v e \tan (A - B) = 1 \Rightarrow \tan (A - B) = \tan \frac{π}{4} \Rightarrow A - B = n π + \frac{π}{4}$

$\Rightarrow A - B = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}, . . . .$

$A l s o \sec (A + B) = \frac{2}{\sqrt{3}} \Rightarrow \cos (A + B) = \frac{\sqrt{3}}{2} = \cos \frac{π}{6} \Rightarrow A + B = 2 n π \pm \frac{π}{6} = \frac{π}{6}, \frac{11 π}{6}, \frac{13 π}{6} . . . .$

$S i n c e A a n d B a r e t o b e t h e l e a s t p o s i t i v e v a l u e s, w e m u s t h a v e A - B = \frac{π}{4} a n d A + B = \frac{11 π}{6} \Rightarrow 2 A = \frac{50 π}{24} \Rightarrow A = \frac{25 π}{24} A l s o 2 B = \frac{38 π}{24} \Rightarrow B = \frac{19 π}{24}$