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In Delta ABC, "cot"(A)/(2) + "cot" (B)/(...

In `Delta ABC, "cot"(A)/(2) + "cot" (B)/(2) + "cot" (C)/(2)` is equal to

A

`(Delta)/(r^(2))`

B

`((a + b + c)^(2))/(abc) 2R`

C

`(Delta )/(r)`

D

`(Delta)/(Rr)`

Text Solution

Verified by Experts

The correct Answer is:
A
`cot.(A)/(2) + cot.(B)/(2) + cot.(C)/(2)`
`= (s(s-a))/(Delta) + (s(s-b))/(Delta) + (s-(s -c))/(Delta)`
`= (s)/(Delta) [3s - (a + b + c)] = (s^(2))/(Delta) = ((Delta//r)^(2))/(Delta) = (Delta)/(r^(2))`
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