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Question

In any ΔABC, prove that

(b2c2) cot A+(c2a2) cot B+(A2B2) cot C=0

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Solution

By the sine rule, we have

asin A=bsin B=csin C

sin Aa=sin Bb=sin Cc=k(say)

sin a=ka,sin b=kb and sin C=kc.

(b2c2)cot A=(b2c2).cos Asin A=(b2c2)ka.(b2+c2a2)2bc

= 1(2babc).(b2c3)(b2+c2a2)=b4c4a2b2+a2c22kabc

Similiary, (c2a2)cot B=(c4a4b2c2+a2b2)2kabc.

And, (a2b2)cot C=(a4b4a2c2+b2c2)2kabc.

LHS=(b2c2)cot A+(c2a2)cot B+(a2b2)cot C

= 12kabc.[(b4c4a2b2+a2c2)+(c4a2b2c2+a2b2)+(a4b4a2c2+b2c2)]

=12kabc×0=0=RHS

Hence, (b2c2)cotA+(c2a2)cot B+(a2b2)cot C=0


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