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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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[(cosx-cosy)/(sinx+siny)] + [(sinx-siny)/(cosx+cosy)] = [cos^2x – cos^2 y + sin^2x – sin^2y]/[(sinx + siny)(cosx + cosy)] = [1-1]/[(sinx + siny)(cosx + cosy)] = 0 ——– Attn: Since [(cosx-cosy)/(sinx+siny)] + [(sinx-siny)/(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