# If CosA = mCosB, then prove that :-No rush for points, please answer with good explanations.

1
by Reeshabh

2014-06-28T13:21:54+05:30

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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 ?
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thankx