in any triangle abc prove that …
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Sia ? 6 years, 6 months ago
L.H.S. = {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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