Question

The parameter, on which the value of the determinant $\left|\begin{array}{ccc}1& a& a^2\\ \cos \left(\left(p-d\right)x\right)& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ \sin \left(\left(p-d\right)x\right)& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right|$ does not depend upon is

• A. $a$
• B. $p$
• C. $d$
• D. $x$

Answer 1

The correct option is B

$p$

Explanation for correct option

Let,

$△=$$\left|\begin{array}{ccc}1& a& a^2\\ \cos \left(\left(p-d\right)x\right)& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ \sin \left(\left(p-d\right)x\right)& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right|$

Now

$C_1\to C_1+C_3$

$$\begin{gathered} \Rightarrow △=\left|\begin{array}{ccc}1+a^2& a& a^2\\ \cos \left(\left(p-d\right)x\right)+\cos \left(\left(p+d\right)x\right)& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ \sin \left(\left(p-d\right)x\right)+\sin \left(\left(p+d\right)x\right)& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right| \\ \Rightarrow △=\left|\begin{array}{ccc}1+a^2& a& a^2\\ 2\cos \left(px\right)\cos \left(dx\right)& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ 2\sin \left(px\right)\cos \left(dx\right)& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right|\left[\because \cos \left(a+b\right)+\cos \left(a-b\right)=2\cos \left(a\right)\cos \left(b\right),\sin \left(a+b\right)+\sin \left(a-b\right)=2\sin \left(a\right)\cos \left(b\right)\right] \end{gathered}$$

$C_1\to C_1-2\cos \left(dx\right)C_2$

$$\begin{gathered} \Rightarrow △=\left|\begin{array}{ccc}1+a^2-2a\cos \left(dx\right)& a& a^2\\ 2\cos \left(px\right)\cos \left(dx\right)-2\cos \left(px\right)\cos \left(dx\right)& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ 2\sin \left(px\right)\cos \left(dx\right)-2\sin \left(px\right)\cos \left(dx\right)& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right| \\ \Rightarrow △=\left|\begin{array}{ccc}1+a^2-2a\cos \left(dx\right)& a& a^2\\ 0& \cos \left(px\right)& \cos \left(\left(p+d\right)x\right)\\ 0& \sin \left(px\right)& \sin \left(\left(p+d\right)x\right)\end{array}\right| \end{gathered}$$

Now expanding with respect to first column

$$\begin{gathered} \Rightarrow △=\left(1+a^2-2a\cos \left(dx\right)\right)\left[\sin \left(\left(p+d\right)x\right)\cos \left(px\right)-\cos \left(\left(p+d\right)x\right)\sin \left(px\right)\right] \\ \Rightarrow △=\left(1+a^2-2a\cos \left(dx\right)\right)\sin \left(dx\right)\because \left[\sin \left(a-b\right)=\sin \left(a\right)\cos \left(b\right)-\cos \left(a\right)\sin \left(b\right)\right] \end{gathered}$$

There is no term related to $p$

Therefore the determinant is independent of $p$

Hence, the correct option is $\left(\mathrm{B}\right)$.