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Question

In a cyclic quadrilateral ABCD, Prove that
tan2B2=(sa)(sb)(sc)(sd).

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Solution

We have
cosB=a2+b2c2d22(ab+cd)
tan2B2=1cosB1+cosB
2(ab+cd)(a2+b2c2d2)2(ab+cd)(a2+b2c2d2)
=(c+d)2(ab)2(a+b)2(cd)2=(c+d+ab)(c+da+b)(a+b+cd)(a+bc+d)
=(2s2b)(2s2a)(2s2d)(2s2c)=(sa)(sb)(sc)(sd)

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