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Given b = a tan B and c = a sec B

Arc tan (a/(b+c))

= arc tan (1/(tan B+sec B))

= arc tan (Cos B /(Sin B + 1))

= arc tan [ (Cos² B/2 - Sin² B/2) / (2 sin B/2 Cos B/2 + SIn² B/2 + Cos² B/2)]

= arc tan [ (Cos B/2 - Sin B/2) / (Cos B/2 + Sin B/2) ]

= arc tan [ (1 - tan B/2) / (1 + tan B/2) ]

= arc tan [ tan (π/4 - B/2) ]

= π/4 - B/2

Similarly,

c = b Sec A and a = b Tan A

Hence, arc tan (b/(c+a)) = π/4 - A/2 following the above.

the answer is

π/2 - (A+B)/2 = π/2 - π/4

*= π/4*

Arc tan (a/(b+c))

= arc tan (1/(tan B+sec B))

= arc tan (Cos B /(Sin B + 1))

= arc tan [ (Cos² B/2 - Sin² B/2) / (2 sin B/2 Cos B/2 + SIn² B/2 + Cos² B/2)]

= arc tan [ (Cos B/2 - Sin B/2) / (Cos B/2 + Sin B/2) ]

= arc tan [ (1 - tan B/2) / (1 + tan B/2) ]

= arc tan [ tan (π/4 - B/2) ]

= π/4 - B/2

Similarly,

c = b Sec A and a = b Tan A

Hence, arc tan (b/(c+a)) = π/4 - A/2 following the above.

the answer is

π/2 - (A+B)/2 = π/2 - π/4