Question

Let A, B, C, D be positive angles such that A + B + C + D = 180°, show that sinA. sinB+ sinC. sinD = sin (B + C) . sin (B + D)

Answer 1

hii mate ☺️

Taking L.H.S

sinA.sinB+sinC.sinD = > 2sinA.sinB+ 2sinC.sinD 

=> cos(A-B) – cos(A+B) +  cos(C-D) – cos(C+D) 

-2sin(A-B+A+B)/2sin(A-B-A-B)/2  +  2sin(C-D+C+D)/2sin(C-D-C-D)/2

=> -2sinA

hope it helps you!!!!!!