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Question

If in triangle ABC,∣ ∣abcbcacab∣ ∣=0 then cotA+cotB+cotC=

A
33
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B
3
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C
13
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D
133
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Solution

The correct option is B 3
∣ ∣abcbcacab∣ ∣

R1R1+R2+R3

a+b+c∣ ∣111bcacab∣ ∣

1(bca2)(b2ac)+abc2=0

212(a2+b2+c2abbcac)=0

(ab)2+(bc)2(ca)2=0

ab=0

bc=0

ca=0

a=b=c.

cotA=cotB=cotC=cot(60)

according to the qs we have 3cotA

=33

=3

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