if 3tanAtanB=1,then prove that 2cos(A+B)=cos(A-B)

Given: {tex}3\tan {\rm{A}}\tan {\rm{B}} = 1{/tex}

=>     {tex}{{3\sin {\rm{A}}\sin {\rm{B}}} \over {\cos {\rm{A}}\cos {\rm{B}}}} = 1{/tex}

=>     {tex}3\sin {\rm{A}}\sin {\rm{B}} = \cos {\rm{A}}\cos {\rm{B}}{/tex}

=>     {tex}3\left[ {{1 \over 2}\left\{ {\cos \left( {{\rm{A}} - {\rm{B}}} \right) - \cos \left( {{\rm{A}} + {\rm{B}}} \right)} \right\}} \right] = {1 \over 2}\left[ {\cos \left( {{\rm{A}} - {\rm{B}}} \right) + \cos \left( {{\rm{A}} + {\rm{B}}} \right)} \right]{/tex}

=>     {tex}3\cos \left( {{\rm{A}} - {\rm{B}}} \right) - 3\cos \left( {{\rm{A}} + {\rm{B}}} \right) = \cos \left( {{\rm{A}} - {\rm{B}}} \right) + \cos \left( {{\rm{A}} + {\rm{B}}} \right){/tex}

=>     {tex} - 4\cos \left( {{\rm{A}} + {\rm{B}}} \right) = - 2\cos \left( {{\rm{A}} - {\rm{B}}} \right){/tex}

=>     {tex}2\cos \left( {{\rm{A}} + {\rm{B}}} \right) = \cos \left( {{\rm{A}} - {\rm{B}}} \right){/tex}

Hence proved.