Question

if tany=Qsinx/P+Qcosx, prove that tan (x-y)=Psinx/Q+Pcosx.

Answer 1

tany=Qsinx/(P+Qcosx)

=>tan{x-(x-y)} =Qsinx /(P+Qcosx)

=>{tanx -tan(x-y)}/{1+tanx.tan(x-y)}=Qsinx/(P+Qcosx)

=>{tanx-tan(x-y)}(P+Qcosx)=Qsinx{1+tanx.tan(x-y)}

=>{sinx-tan(x-y).cosx}(P+Qcosx)=Qsinx{cosx +sinx.tan(x-y)}

=>(P+Qcosx)sinx-tan(x-y)(P+Qcosx)cosx=
Qsinx.cosx +Qsin^2x.tan(x-y)

=>-tan(x-y).Pcosx+(P+Qcosx)sinx=
Qsinx.cosx +Qtan(x-y)

=>sinx.P =tan(x-y){Q+Pcosx}

=>tan(x-y)=Psinx/(Q+Pcosx)
hence proved