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  • In triangle ABC  <img alt="\frac{4b^2c^2-(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{8b^2c^2}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B4b%5E2c%5E2-%28a&plus;b&plus;c%29%28b&amp

In triangle ABC  \frac{4b^2c^2-(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{8b^2c^2} is equal to ?

Option: 1

\sin^2 A


Option: 2

\cos^2 A


Option: 3

\frac{1}{2} \sin^2 A


Option: 4

\frac{1}{2} \cos^2 A


Answers (1)

best_answer

\\\frac{4b^2c^2-(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{8b^2c^2}\\ =\frac{1}{2}-\frac{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}{8b^2c^2}\\ =\frac{1}{2}-\frac{2 s \times 2\times (s-a)\times 2\times (s-b)\times 2\times (s-c)}{8b^2c^2}\\ =\frac{1}{2}-\frac{ 2s \times (s-a)\times (s-b)\times (s-c)}{b^2c^2}\\ =\frac{1}{2}-\frac{ 2\Delta^2 }{b^2c^2}\\\\ =\frac{1}{2}-\frac{ 2 (\frac{1}{2}bc \sin A)^2 }{b^2c^2}\\\\ =\frac{1}{2} \cos^2 A

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shivangi.shekhar

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