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Question

Whether the given equation (b2c2a2)sin2A+(c2a2b2)sin2B+(a2b2c2)sin2C=1 ?(In any ΔABC )

A
True
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B
False
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Solution

The correct option is B False
Prove that (b2c2)a2sin2A+(c2a2)b2sin2B+(a2b2)c2sin2C=1
By sine rule sinAa=sinBb=sinCc=k
cosA=b2+c2a22bc
LHS
(b2c2a2)2ak+(b2+c2a22bc)+(c2a2b2)2bk(a2+c2b22ac)+(a2b2c2)2ck(a2+b2c22ab)
=kabc(b2c2)(b2+c2a2)+(c2a2)(a2+c2b2)+(a2b2)(a2+b2c2)
=kabc(b2b2/c2a2/b2a2c2c4+a2c2+a2c2+c4b2/c2a4a2c2+a2b2+a4+a2b2a2c2a2b2b4+b2c2)
=kabc(0)
=0
Hence 0=1
LHS RHS
Hence not proved.

1355244_1132447_ans_0a1e8b3d17224028a003e2b7ba29ba90.jpg

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