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What is the value of \(\begin{vmatrix}1 & cosx-sinx & cosx + sinx \\1 & cosy-siny & cosy + siny \\1 & cosz-sinz & cosz + sinz \end {vmatrix}\)?(a) 3\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)(b) \(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)(c) 2\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)(d) 4\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)The question was asked in class test.My doubt stems from Determinant topic in chapter Determinants of Mathematics – Class 12

Answer»

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\(\begin{vmatrix}1 & cosx & sinx \\1 & cosy & siny \\1 & cosz & sinz \end {vmatrix}\)


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