In triangle ABC, if `|[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B/2)+tan(C/2), tan(C/2)+ tan(A/2), tan(A/2)+ tan(B/2)]|` =0 then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of these

Options

A. equilateral
B. isosceles
C. obtuse angled
D. none of these

Answer

B

Explanation

Applying `C_(1) to C_(1)-C_(2),C_(2) to C_(2)-C_(3)` we get
`|{:(0,,0,,1),(cot .(A)/(2)-cot.(B)/(2),,cot.(B)/(2)-cot.(C)/(2),,cot.(C)/(2)),(tan.(B)/(2)-tan.(A)/(2),,tan.(C)/(2)-tan.(B)/(2),,tan.(A)/(2)+tan.(B)/(2)):}|`
`|{:(0,,0,,1),(cot.(A)/(2)-cot.(B)/(2),,cot.(B)/(2)-cot.(C )/(2),,cot.(C)/(2)),((cot.(A)/(2)-cot.(B)/(2))/(cot.(A)/(2)-cot.(A)/(2)),,(cot.(B)/(2)-cot.(C)/(2))/(cot.(B)/(2)cot.(C)/(2)),,tan.(A)/(2)+tan.(B)/(2)):}|`
`=( cot .(A)/(2)-cot.(B)/(2))(cot.(B)/(2)-cot.(C)/(2))`
`xx |{:(0,,0,,0),(1,,1,,cot.(C)/(2)),(tan.(A)/(2)-cot.(B)/(2),,tan.(B)/(2)tan.(C)/(2),,tan.(A)/(2).tan.(B)/(2)):}|`
`=(cot.(A)/(2)-cot.(B)/(2))(cot.(B)/(2)-cot.(C)/(2))(tan.(C)/(2)-tan.(A)/(2))tan.(B)/(2)`
Since `Delta =0 ` we have
:. `cot .(A)/(2)=cot.(B)/(2) " or " cor .(B)/(2) -cot .(C)/(2)" or " tan.(A)/(2) =tan .(C)/(2)`
Hence the triangle is definitely isosceles.