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solve for left side to equal 0
4 Answers

(cosx  cosy / sinx + siny) + (sinx  siny / cosx + cosy) = 0 =====> LCD
[ (cosx  cosy) * (cosx + cosy) / (sinx + siny) * (cosx + cosy) + [ (sinx  siny)*(sinx + siny) / (cosx + cosy) * (sinx + siny) ] = 0
[ ( cos^2(x) + cos(x)cos(y)  cos(y)cos(x)  cos^2(y) ) / (sinx + siny) * (cosx + cosy) + [ (sin^2(x) + sin(x)sin(y)  sin(y)sin(x)  sin^2(y)) / (cosx + cosy) * (sinx + siny) ]
[ ( cos^2(x) + cos(x)cos(y)  cos(y)cos(x)  cos^2(y) ) + ( sin^2(x) + sin(x)sin(y)  sin(y)sin(x)  sin^2(y) ) ] / [ (cosx + cosy) * (sinx + siny) ]
[ cos^2(x)  cos^2(y) + sin^2(x)  sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
[ cos^2(x) + sin^2(x)  cos^2(y)  sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
knowing the identities;
cos^2(x) + sin^2(x) = 1
[ 1  cos^2(y)  sin^2(y) ] / [ (cosx + cosy) * (sinx + siny) ]
[ 1  ( cos^2(y) + sin^2(y) ) ] / [ (cosx + cosy) * (sinx + siny) ]
[ 1  ( 1 ) ] / [ (cosx + cosy) * (sinx + siny) ]
[ 1  1 ] / [ (cosx + cosy) * (sinx + siny) ]
[ 0 ] / [ (cosx + cosy) * (sinx + siny) ]
0

Sinx Siny
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[(cosxcosy)/(sinx+siny)] + [(sinxsiny)/(cosx+cosy)] = [cos^2x  cos^2 y + sin^2x  sin^2y]/[(sinx + siny)(cosx + cosy)] = [11]/[(sinx + siny)(cosx + cosy)] = 0  Attn: Since [(cosxcosy)/(sinx+siny)] + [(sinxsiny)/(cosx+cosy)] is given, that means it is well defined and (sinx+siny)(cosx+cosy) not = 0

(cosx  cosy / sinx + siny) + (sinx  siny / cosx + cosy)
= [(cos^2x  cos^2y) + (sin^2x  sin^2y)] / [(sinx + siny)(cosx + cosy)]
= [cos^2x+sin^2x  (cos^2y+sin^2y)] / [(sinx + siny)(cosx + cosy)]
= (1  1)/ [(sinx + siny)(cosx + cosy)]
= 0