Question

If in Δ ABC, 3a = b + c, then `cot ("B"/2) cot ("C"/2)` = ______.

Options

• 1

• 2

• 4

• 5

Answer 1

If in Δ ABC, 3a = b + c, then `cot ("B"/2) cot ("C"/2)` = 2.

Explanation:

`cot ("B"/2) * cot ("C"/2) = sqrt(("s"("s" - "b"))/(("s - a")("s - c"))) * sqrt(("s"("s" - "c"))/(("s - a")("s - b")))`

`= "s"/("s - a")`    ....(i)

But 3a = b + c

⇒ 3a + a = a + b + c

⇒ 4a = a + b + c

⇒ s = 2a

Putting s = 2a in (i), we get

`cot ("B"/2) * cot ("C"/2) = "2a"/(2"a" - "a")`

`= "2a"/"a" = 2`