Question

If ΔABC is right angled at C then arc tan(a/(b+c))+ arc tan(b/(c+a))=

Answer 1

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