21. (b+c-a) (cor
+ cot) = 2a cot A
A
2.
Answers
Answer:
a=k sin A
b=k sin B
c=k sin C
(b+c-a)(cot B/2+cot C/2)=2 a cot A/2
(b+c-a)/a=(k sin B+k sin C-k sin A)/k sin A
=k(sin B+sin C-sin A)/k sin A
=(sin B+sin C-sin A)/sin A
sin C+sin D=2 sin (C+D)/2.cos(C-D)/2 AND sin 2 A=2 sin A . cos A
A+B+C=180;
Is (b+c-a) (cot B/2+cot C/2) =2a cot A/2?
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2 Answers

Subandi Jayaweera, studied at Dharmasoka College
Updated July 17
a/Sin A=b/Sin B=c/Sin C=k ( SIN THEOREM)
so, a=k sin A
b=k sin B
c=k sin C
(b+c-a)(cot B/2+cot C/2)=2 a cot A/2
(b+c-a)/a=(k sin B+k sin C-k sin A)/k sin A
=k(sin B+sin C-sin A)/k sin A
=(sin B+sin C-sin A)/sin A
sin C+sin D=2 sin (C+D)/2.cos(C-D)/2 AND sin 2 A=2 sin A . cos A
A+B+C=180;
so,
(sin B+sin C-sin A)/sin A=[2sin (B+C)/2.cos (B-C)/2– 2sin A/2.cosA/2]/2sin A/2.cosA/2
(B+C)/2=(180-A)/2;
=[sin(90-A/2).cos(B-C)/2-sin A/2.cos A/2]/sin A/2.cos A/2
sin (90-A)=cos A;
=[cos A/2.cos(B-C)/2-sin A/2.cos A/2]/sin A/2.cos A/2
=cos A/2[cos(B-C)/2-sin A/2]/sin A/2.cos A/2
(cos A)/(sin A)=cot A;
={cot A/2[cos(B-C)/2-sin A/2]}/cos A/2
A/2=[180-(B+C)]/2;
=cot A/2 [cos (B-C)/2-sin 90–(B+C)/2]/cos A/2