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)]|` then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of theseA. equilateralB. isoscelesC. obtuse angledD. none of these

In triangle ABC, if `|[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B

1.	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)]\|` then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of theseA. equilateralB. isoscelesC. obtuse angledD. none of these
Answer» Correct Answer - B 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.