Question

sinA+ sinB+ sinC/sinA +sinB-sinC=cotA/2 cotB/2​

Answer 1

Answer:

Let A+B+C=θ

sin(A+B2)=cos(C2)sin(A+B2)=cos(C2)

D→sinA+sinB−sinCD→sinA+sinB-sinC

=2sin(A+B2)cos(A−B2)−sinC=2sin(A+B2)cos(A-B2)-sinC

=2cos(C2)cos(A−B2)−2sin(C2)cos(C2)=2cos(C2)cos(A-B2)-2sin(C2)cos(C2)

=2cos(C2)[cos(A−B2)−sin(C2)]=2cos(C2)[cos(A-B2)-sin(C2)]

=2cos(C2)[cos(A−B2)−cos(A+B2)]=2cos(C2)[cos(A-B2)-cos(A+B2)]

=2cos(C2)[2sin(A2)sin(B2)]=2cos(C2)[2sin(A2)sin(B2)]

=4sinn(A2)sin(B2)cos(C2)=4sinn(A2)sin(B2)cos(C2)

N→sinA+sinB+sinCN→sinA+sinB+sinC

=2sin(A+B2)cos(A−B2)+2sin(C2)cos(C2)=2sin(A+B2)cos(A-B2)+2sin(C2)cos(C2)

=2cos(C2)cos(A−B2)+2sin(C2)cos(C2)=2cos(C2)cos(A-B2)+2sin(C2)cos(C2)

=2cos(C2)[cos(A−B2)+cos(A+B2)]=2cos(C2)[cos(A-B2)+cos(A+B2)]

=2cos(C2)[2cos(A2)cos(B2)]=2cos(C2)[2cos(A2)cos(B2)]

=4cos(A2)cos(B2)cos(C2)=4cos(A2)cos(B2)cos(C2)

ND=4cos(A2)cos(B2)cos(C2)4sin(A2)sin(B2)cos(C2)ND=4cos(A2)cos(B2)cos(C2)4sin(A2)sin(B2)cos(C2)

=cot(A2)cot(B2)=cot(A2)cot(B2).

Hence,Proved..