If cotθ=cos(x+y) and cotΦ=cos(x-y) show that tan(θ-Φ)=2.sinx.siny/(cos^2x+cos^2y)

Steps to prove tan(θ-Φ)=2.sinx.siny/(cos^2x+cos^2y)

L.H.S

=tan(θ-Φ)

=(tanθ-tanΦ)/(1-tanθ.tanΦ)

=(1/cotθ-1/cotΦ)/(1-1/cotθ.1/cotΦ)

=(cotΦ-cotθ)/(cotθ.cotΦ+1)

={cos(x-y)-cos(x+y)}/{cos(x-y).cos(x+y)+1}

=2sinx.siny/(cos^2x-sin^2y+1)

=2sinx.siny/(cos^2x+1-sin^2y)

=2.sinx.siny/(cos^2x+cos^2y)

R.H.S

Hence tan(θ-Φ)=2.sinx.siny/(cos^2x+cos^2y) is proved.

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