Right option is (c) 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)
For explanation I would say: Let, a = cosx, b = cosy, c = cosz, p =sinx, q = siny and r = sinz
So, \(\begin{vmatrix}1 & a – p & a + p \\1 & b – q & b + q \\1 & c – r & c + r \end {vmatrix}\)
Making C3 = C3 + C2
= \(\begin{vmatrix}1 & a – p & 2a \\1 & b – q & 2b \\1 & c – r & 2c \end {vmatrix}\)
= 2\(\begin{vmatrix}1 & a – p & a \\1 & b – q & b \\1 & c – r & c \end {vmatrix}\)
Making C2 = C2 – C3
= -2\(\begin{vmatrix}1 & p & a \\1 & q & b \\1 & r & c \end {vmatrix}\)
Interchanging 2^nd and 3^rd column, we get,
2\(\begin{vmatrix}1 & a & p \\1 & b & q \\1 & c & r \end {vmatrix}\)
= 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)