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Question

Let A+B+C=π2 and cotA+cotB+cotC=kcotA.cotB.cotC then the value of k

A
1
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B
12
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C
0
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D
2
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Solution

The correct option is A 1
Given A+B+C=π2
or, A+B=π2C
Then cot(A+B)=tanC
or, cotA.cotB1cotB+cotC=1cotC
or, cotA.cotB.cotCcotC=cotA+cotB
or, cotA+cotB+cotC=cotA.cotB.cotC.
Comparing this with the given equation we get, k=1.

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