Question

$\\\frac{sin(B+C)+sin(C+A)+sin(A+B}{SinA+sinB+sinC} =-1$ Prove this , if A,B,C are the angles of a Triangle

Answer 1

A,B,C angles of triangle.

. ' .A+B+C = 180°

=> B +C =180°-A

=> sin( B +C) = sin(180°- A)

=>sin(B+C)=sinA

sin(C+A)= sin(180°-B)

=> sin( C + A)=sinB

sin(A +B)= sin(180°-C)

=> sin(A + B)= sinC

. ' .L.H.S= sin(B+C)+sin(C+A)+sin(A+B)

sinA + sinB + sinC

= sinA + sinB + sinC

sinA + sinB+ sinC

= 1.