In ΔABC,∑sin(A+B).sin(A−B)/cos^2 A ...
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In ΔABC, ∑frac{sin(A+B).sin(A-B)}{cos^{2}Acos^{2}B } 1) 0 2) 1 3) 2 4) 1/2

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Maths
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In ΔABC,∑sin(A+B).sin(A−B)/cos^2 A cos^2 B
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∑frac{sin(A+B).sin(A-B)}{cos^{2}Acos^{2}B } =∑frac{sin^{2}Acos^{2}B-cos^{2}Asin^{2}B}{cos^{2}Acos^{2}B} =∑(tan^{2}A-tan^{2}B)=(tan^{2}A-tan^{2}B)+(tan^{2}B-tan^{2}C)+(tan^{2}C-tan^{2}A)=0

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