How to make this expression independent from B? thank you!

Question

#(sin(B- A) + cos A + sin(B + A))/((2 tan B + sec B) *cos B#

I don't whether that's the right transltion. But the goal of this exercise is to make sure you have a B in your final answer anymore using Simpsons formula's, double-angle formula's, sum and difference formula's... It's for a preparation for an exam, but I can't seem to figure it out. Thank you!

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Answer 1 · 2017-12-08T19:14:18

Please refer to the Discussion given in the Explanation.

We will use : #sinC+sinD=2sin((C+D)/2)cos((C-D)/2).#

#"The Expression="{sin(B-A)+cosA+sin(B+A)}/{(2tanB+secB)cosB},#

#=[{sin(B+A)+sin(B-A)}+cosA]/{(2tanB+secB)cosB},#

#=[{2sinBcosA}+cosA]/{(2tanB+secB)cosB},#

#=[{2sinBcosA}+cosA]/{2tanBcosB+secBcosB},#

#=[{2sinBcosA}+cosA]/{(2sinB)/cosB*cosB+1/cosB*cosB},#

#=[{2sinBcosA}+cosA]/(2sinB+1),#

#=[cancel((2sinB+1))cosA}/cancel((2sinB+1)),#

#=cosA,# which is free from #B.#

Q.E.D.

Answer 2 · 2017-12-08T19:59:25

See below.

We have

#g(a,b) = u(a)v(b)# then

#{(g(0,b)=u(0)v(b)),(g(a,0)=u(a)v(0)),(g(0,0)=u(0)v(0)):}#

and then

#g(a,b) = (g(0,b)g(a,0))/g(0,0)#

but now with #a = A# and #b = B#

#g(0,B) =(SecB (1 + 2 SinB))/(SecB + 2 TanB)#
#g(A,0) = cosA#
#g(0,0) = 1#

and with those considerations we conclude:

#(Sin(B-A) + CosA + Sin(B-A))/((2 TanB + SecB) CosB) = v(B)cos(A)#

but #v(B) = 1#

then finally

#(Sin(B-A) + CosA + Sin(B-A))/((2 TanB + SecB) CosB) = cos(A)#