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A+B+C =  tan(A/2 + B/2) = tan( /2 - C/2) = cotC/2Arranging it we get tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2 = 1 Now tan2A/2 + tan2B/2 + tan2C/2 - 1 = tan2A/2 + tan2B/2 + tan2C/2 - (tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2) = 1/2[(tanA/2-tanB/2)2 + (tanB/2-tanC/2)2 + (tanC/2-tanA/2)2 >= 0 Hence tan2A/2 + tan2B/2 + tan2C/2 - 1 >= 0 tan2A/2 + tan2B/2 + tan2C/2 >= 1
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Bipin Kumar Dubey Chemical Dept. IIT Kharagpur |
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