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

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

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