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cot12^(@)cot102^(@)+cot102^(@)cot66^(@)+...

`cot12^(@)cot102^(@)+cot102^(@)cot66^(@)+cot66^(@)cot12^(@)` =

A

1

B

`-2`

C

2

D

`-1`

Text Solution

Verified by Experts

The correct Answer is:
A
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