How do you verify the identity (cosx - cosy)/(sinx + siny) + (sinx - siny)/(cosx + cosy) = 0?

We have to prove (cosx – cosy / sinx + siny) + (sinx – siny / cosx + cosy) = 0

Solution

Let us start with LHS

(cosx – cosy / sinx + siny) + (sinx – siny / cosx + cosy)

=[ (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) ]

We know the identity

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

= RHS

Hence Proved

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