If `p=sin(A-B)sin(C-D),q=sin(B-C)sin(A-D)`, `r=sin(C-A)sin(B-D)` then (a) `p+q-r=0` (b) `p+q+r=0` `p-q+r=0` (d) `p^3+q^3+r^3=3p q r`

Explanation

Here, we will use,

`2sinCsinD = cos(C-D) - cos(C+D)`

Now, using the above identity,

`2p = cos(A-B-C+D) - cos(A-B+C-D)->(1)`

`2q = cos(B-C-A+D) - cos(B-C+A-D)->(2)`

`2r = cos(C-A-B+D) - cos(C-A+B-D)->(3)`

Now, adding (1),(2) and (3),

`2(p+q+r) = cos(A-B-C+D) - cos(A-B+C-D)+cos(B-C-A+D) - cos(B-C+A-D)+cos(C-A-B+D) - cos(C-A+B-D)`

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