in any triangle abc prove that (a-b)^2 cos^2c/2 + (a+b)^2 sin^c/2=c^2

1.	in any triangle abc prove that (a-b)^2 cos^2c/2 + (a+b)^2 sin^c/2=c^2
Answer» L.H.S. =\xa0{tex}{(a - b)^2}{\\cos ^2}\\frac{c}{2} + {(a + b)^2}{\\sin ^2}\\frac{c}{2}{/tex}{tex} = ({a^2} + {b^2} - 2ab){\\cos ^2}\\frac{c}{2}{/tex}{tex} + ({a^2} + {b^2} + 2ab){\\sin ^2}\\frac{c}{2}{/tex}{tex} = {a^2}{\\cos ^2}\\frac{c}{2} + {b^2}{\\cos ^2}\\frac{c}{2} - 2ab\\;{\\cos ^2}\\frac{c}{2}{/tex}{tex} + {a^2}{\\sin ^2}\\frac{c}{2} + {b^2}{\\sin ^2}\\frac{c}{2} + 2ab\\;{\\sin ^2}\\frac{c}{2}{/tex}{tex} = {a^2}\\left( {{{\\cos }^2}\\frac{c}{2} + {{\\sin }^2}\\frac{c}{2}} \\right){/tex}{tex} + {b^2}\\left( {{{\\cos }^2}\\frac{c}{2} + {{\\sin }^2}\\frac{c}{2}} \\right){/tex}{tex} = 2ab\\left( {{{\\cos }^2}\\frac{c}{2} - {{\\sin }^2}\\frac{c}{2}} \\right){/tex}= a2 + b2 - 2 ab cosC{tex} = {a^2} + {b^2} - 2ab\\left( {\\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \\right){/tex}= a2 + b2 - a2 - b2 + c2=c2 R.H.S.

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