In any triangles ABC, establish Napier’s formula, `tan((B-C)/2)=(b-c)/(b+c)cot(A/2)`

Choose the correct option In a triangle ABC, sin((A+B)/2) is equal to a) cos(B-C)/2 b) cos(C-A)/2 c) cos(A/2) d) cos (C/2)

Show that in a triangle ABC, (asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)

If A,B and C are interior angle of a triangle ABC,then show that sin(B+C)/2=cos(A/2)

If A,B,C are the three angles of a triangle,show that:(iv) tan((B+C)/2)=cot(A/2)

Show that in a triangle ABC, asin(A/2)sin((B-C)/2)+bsin(B/2)sin((C-A)/2)+csin(C/2)sin((A-B)/2)=0

In a triangle ABC, prove that (a^2sin(B-C))/sinA+(b^2sin(C-A))/sinB+(c^2sin(A-B))/sinC=0

If A,B,C are the interior angles of ABC, then prove that cos((A+B)/2)=sin(C/2) .

In a ∆ ABC prove that tan( (A+B)/2) = cot( C/2) .

If A,B,C are the three angles of a triangle,show that:(iii) sin((B+C)/2)=cos(A/2)