Question

Pls answer quickly!
Show that (cos ^3 α + sin ^3 α)/(cos α + sin α) + (cos ^3 α − sin ^3 α)/(cos α − sin α) = 2

Answer 1

Answer

Hy mate here is your answer

we have

we havecosα+cosβ=a

we havecosα+cosβ=asinα+sinβ=b

we havecosα+cosβ=asinα+sinβ=bα−β=2θ

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)2+cos(α−β)=a2+b2    [sin2α+cos2α=1sin2β+cos2β=1cosαcosβ+sinαsinβ=cos(α−β)

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)2+cos(α−β)=a2+b2    [sin2α+cos2α=1sin2β+cos2β=1cosαcosβ+sinαsinβ=cos(α−β)a2+b2=2+2cos(α−β)

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)2+cos(α−β)=a2+b2    [sin2α+cos2α=1sin2β+cos2β=1cosαcosβ+sinαsinβ=cos(α−β)a2+b2=2+2cos(α−β)=2+2cos2θ

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)2+cos(α−β)=a2+b2    [sin2α+cos2α=1sin2β+cos2β=1cosαcosβ+sinαsinβ=cos(α−β)a2+b2=2+2cos(α−β)=2+2cos2θcosθcos3θ=cosθ4cos3θ.3cosθ

we havecosα+cosβ=asinα+sinβ=bα−β=2θso, (cosα+cosβ)2=a2cos2α+cos2β+2cosαcosβ=a2-----(i)and (sinα+sinβ)2=b2sin2α+sin2β+2sinαsinβ=b2-----(ii)adding (i) and (ii)2+cos(α−β)=a2+b2    [sin2α+cos2α=1sin2β+cos2β=1cosαcosβ+sinαsinβ=cos(α−β)a2+b2=2+2cos(α−β)=2+2cos2θcosθcos3θ=cosθ4cos3θ.3cosθ=4cos

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