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

Given:tany=QsinxP+Q cosxtanx-x-y=QsinxP+Q cosxtanx-tanx-y1+tanx tanx-y=QsinxP+Q cosxsinx-tanx-ycosxP+Q cosx=Qsinxcosx+sinx tanx-ysinxP+Q cosx-tanx-yP+Q cosxcosx=Q sinxcosx+Qsin2x tanx-y-tanx-yP+Q cosxcosx-Qsin2x tanx-y=Q sinxcosx-Psinx-Qsinx cosxtanx-y-Pcosx-Qcos2x-Qsin2x=-P sinxtanx-yPcosx+Qcos2x+sin2x=P sinxtanx-yPcosx+Q=P sinx  using: cos2x+sin2x=1tanx-y=PsinxQ+PcosxHence proved.

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