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In an acute angled triangle $\cot A \cot B + \cot C \cot A + \cot B \cot C =$

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detailed solution

Correct option is B

We haveA+B+C=π⇒A+B=π-C⇒cot(A+B)=cotπ-C⇒cot(A+B)=-cotC⇒cotAcotB-1cotB+cotA=-cotC⇒-cotBcotC-cotAcotC=cotAcotB-1⇒cotAcotB+cotBcotC+cotCcotA=1

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In an acute angled triangle cotAcotB+cotCcotA+cotBcotC=

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In an acute angled triangle $\cot A \cot B + \cot C \cot A + \cot B \cot C =$