Question

sin (B-c) cos (a - d) + sin (c - a) cos (B - d) + sin (a - b) cos(c - d) =0
PROVE IT​

Answer 1

Answer:

Here, we will use,

2sinAcosB=sin(A+B)+sin(A-B)

Now,

L.H.S.=sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(a-B)cos(C-D)

=1/2[2sin(B-C)cos(A-D)+2sin(C-A)cos(B-D)+2sin(a-B)cos(C-D)]

=1/2[sin(B-C+A-D)+sin(B-C-A+D)+sin(C-A+B-D)+sin(C-A-B+D)+sin(A-B+C-D)+sin(A-B-C+D)]

=1/2[sin(B-C+A-D)-sin(A-B+C-D)-sin(A-B-C+D)-sin(A+B-C-D)+sin(A-B+C-D)+sin(A-B-C+D)]

=1/2[0]

Pro