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using coponendo dividendo ie if a/b = c/d then (a+b)/(a-b)= (c+d)/(c-d) so (CosA + CosB)/(CosA - CosB)=(m +1)/(m -1) using formulas [2Cos {(A + B)/2}Cos{(B - A)/2}]/[2Sin{(B- A)/2}Sin{(A+ B )/2}]=(m +1)/(m -1) do cross multiplication 2 will be canceled Cos{(A + B)/2}/Sin{(A + B)/2}={(m +1)/(m -1)} × Sin{(B - A)/2}/Cos {(B - A)/2} we know that cosθ/sinθ = cotθ and sinθ/cosθ= tanθ so we get Cot{(A + B)/2}={(m +1)/(m -1)} Tan{(B - A)/2} PROVED

Oh nice, I didn't know there were such formulas too ! Thanks !

http://www.maths.manchester.ac.uk/~cds/internal/tables/trig.pdf Look here, the formula is different. It shows A-B in the RHS, but in your answer, it is B-A, can you tell why ?

and after expanding CosA-CosB, there should be a negative sign on one of the terms of RHS, where is that in your answer ?