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

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