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Question
In a cyclic quadrilateral ABCD, Prove that
tan2B2=(s−a)(s−b)(s−c)(s−d).
tan2B2=(s−a)(s−b)(s−c)(s−d).
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Solution
We have
cosB=a2+b2−c2−d22(ab+cd)
∴tan2B2=1−cosB1+cosB
2(ab+cd)−(a2+b2−c2−d2)2(ab+cd)−(a2+b2−c2−d2)
=(c+d)2−(a−b)2(a+b)2−(c−d)2=(c+d+a−b)(c+d−a+b)(a+b+c−d)(a+b−c+d)
=(2s−2b)(2s−2a)(2s−2d)(2s−2c)=(s−a)(s−b)(s−c)(s−d)
cosB=a2+b2−c2−d22(ab+cd)
∴tan2B2=1−cosB1+cosB
2(ab+cd)−(a2+b2−c2−d2)2(ab+cd)−(a2+b2−c2−d2)
=(c+d)2−(a−b)2(a+b)2−(c−d)2=(c+d+a−b)(c+d−a+b)(a+b+c−d)(a+b−c+d)
=(2s−2b)(2s−2a)(2s−2d)(2s−2c)=(s−a)(s−b)(s−c)(s−d)
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