Question

The parameter on which the value of the determinant $\left| {\begin{array}{*{20}{c}}
  1&a&{{a^2}} \\
  {\cos \left( {p - d} \right)x}&{\cos px}&{\cos \left( {p + d} \right)x} \\
  {\sin \left( {p - d} \right)x}&{\sin px}&{\sin \left( {p + d} \right)x}
\end{array}} \right|$ does not depend upon is
(A) p
(B) d
(C) a
(D) p and a

Answer 1

Hint: Here we will use the properties of determinants to solve the above determinant. We will try to make two zeros in column one by first subtracting it with column three and then subtracting it again with some multiple of column two as per requirement.

Complete step by step solution:
In the above question, we will apply some properties of determinants to solve it.
These properties can be applied to rows as well as columns.
Let $\vartriangle \, = \,\left| {\begin{array}{*{20}{c}}
  1&a&{{a^2}} \\
  {\cos \left( {p - d} \right)x}&{\cos px}&{\cos \left( {p + d} \right)x} \\
  {\sin \left( {p - d} \right)x}&{\sin px}&{\sin \left( {p + d} \right)x}
\end{array}} \right|$
Applying, ${C_1} \to {C_1} + {C_3}$
$ \Rightarrow \vartriangle \, = \,\left| {\begin{array}{*{20}{c}}
  {1 + {a^2}}&a&{{a^2}} \\
  {\cos \left( {p - d} \right)x + \cos \left( {p + d} \right)x}&{\cos px}&{\cos \left( {p + d} \right)x} \\
  {\sin \left( {p - d} \right)x + \sin \left( {p + d} \right)x}&{\sin px}&{\sin \left( {p + d} \right)x}
\end{array}} \right|$
Now, we will use identity $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ in column one.
$ \Rightarrow \vartriangle \, = \,\left| {\begin{array}{*{20}{c}}
  {1 + {a^2}}&a&{{a^2}} \\
  {2\cos \left( {px} \right)\cos \left( {dx} \right)}&{\cos px}&{\cos \left( {p + d} \right)x} \\
  {2\sin px\cos dx}&{\sin px}&{\sin \left( {p + d} \right)x}
\end{array}} \right|$
Now, we will apply ${C_1} \to {C_1} - 2\cos dx\,{C_2}$
$ \Rightarrow \vartriangle \, = \,\left| {\begin{array}{*{20}{c}}
  {1 + {a^2} - 2a\cos dx}&a&{{a^2}} \\
  0&{\cos px}&{\cos \left( {p + d} \right)x} \\
  0&{\sin px}&{\sin \left( {p + d} \right)x}
\end{array}} \right|$
Whenever we get two zeros in a single column or a single row, we will solve that determinant about that row or column.
Here, we are getting two zeros in column one. So, we will solve this determinant using the formula $ \Rightarrow \vartriangle = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)$
$ \Rightarrow \vartriangle \, = \,\left( {1 + {a^2} - 2a\,\cos dx} \right)\left[ {\sin \left( {p + d} \right)x\cos px - \sin px\cos \left( {p + d} \right)x} \right] - 0 + 0$
Now, using the identity $\sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \cos \left( A \right)\sin \left( B \right)$
$ \Rightarrow \vartriangle \, = \,\left( {1 + {a^2} - 2a\,\cos dx} \right)\left[ {\sin \left( {p + d - p} \right)x} \right]$
$ \Rightarrow \vartriangle \, = \,\left( {1 + {a^2} - 2a\,\cos dx} \right)\left[ {\sin \left( d \right)x} \right]$
Here we can see that the above expression is independent of p.

Therefore, the correct option is A

Note: Important Properties of Determinants are:
$\left( 1 \right)$ Reflection Property: The determinant remains unaltered if its rows are changed into columns and the columns into rows. This is known as the property of reflection.
$\left( 2 \right)$ All-zero Property: If all the elements of a row (or column) are zero, then the determinant is zero.
$\left( 3 \right)$ Proportionality (Repetition) Property: If the all elements of a row (or column) are proportional (identical) to the elements of some other row (or column), then the determinant is zero.
$\left( 4 \right)$ Switching Property: The interchange of any two rows (or columns) of the determinant changes its sign.
$\left( 5 \right)$Scalar Multiple Property: If all the elements of a row (or column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.