Question

A value of $\theta \in \left( 0,\pi /3 \right)$, for which $\left| \begin{matrix} 1+{{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$, is
$\text{(A) }\dfrac{7\pi }{24}$
$\text{(B) }\dfrac{\pi }{18}$
$\text{(C) }\dfrac{\pi }{9}$
$\text{(D) }\dfrac{7\pi }{36}$

Answer 1

In this question we are given a determinant which has trigonometric functions in it. We are given that the modulus value of the given determinant is $0$. Now we can use the approach of directly calculating the mod but we will first simplify the determinant using row transformations and then find then equate the value of the mod with $0$ and then find the value of $\theta$ such that it belongs in the range of $\left( 0,\pi /3 \right)$.

Complete step by step solution:
We have the determinant given to us as:
$\Rightarrow \left| \begin{matrix} 1+{{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
Now on doing the transformation ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$
$\Rightarrow \left| \begin{matrix} 1+{{\cos }^{2}}\theta -\left( {{\cos }^{2}}\theta \right) & {{\sin }^{2}}\theta -\left( 1+{{\sin }^{2}}\theta \right) & 4\cos 6\theta -\left( 4\cos 6\theta \right) \\\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
On opening the brackets, we get:
$\Rightarrow \left| \begin{matrix} 1+{{\cos }^{2}}\theta -{{\cos }^{2}}\theta & {{\sin }^{2}}\theta -1-{{\sin }^{2}}\theta & 4\cos 6\theta -4\cos 6\theta \\\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
Now since the same terms with opposite signs cancel each other, we get:
$\Rightarrow \left| \begin{matrix} 1 & -1 & 0 \\\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
Now on doing the transformation ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$
$\Rightarrow \left| \begin{matrix} 1 & -1 & 0 \\\ {{\cos }^{2}}\theta -\left( {{\cos }^{2}}\theta \right) & 1+{{\sin }^{2}}\theta -\left( {{\sin }^{2}}\theta \right) & 4\cos 6\theta -\left( 1+4\cos 6\theta \right) \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
On opening the brackets, we get:
$\Rightarrow \left| \begin{matrix} 1 & -1 & 0 \\\ {{\cos }^{2}}\theta -{{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta -{{\sin }^{2}}\theta & 4\cos 6\theta -1-4\cos 6\theta \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$
Now since the same terms with opposite signs cancel each other, we get:

1 & -1 & 0 \\\ 0 & 1 & -1 \\\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\\ \end{matrix} \right|=0$$ Now on taking the mod of the determinant, we get: $\Rightarrow 1\left( \left( \left( 1 \right)\times \left( 1+4\cos 6\theta \right) \right)-\left( \left( -1 \right)\times \left( {{\sin }^{2}}\theta \right) \right) \right)-\left( -1 \right)\left( 0- \left(-{{\cos }^{2}}\theta\right) \right)+0=0$ On opening the brackets, we get: $\Rightarrow 1\left( 1+4\cos 6\theta -\left( -{{\sin }^{2}}\theta \right) \right)+1\left( {{\cos }^{2}}\theta \right)=0$ On further simplifying, we get: $$\Rightarrow 1+4\cos 6\theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =0$$ Now we know the trigonometric identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ therefore, on using this identity, we get: $\Rightarrow 1+4\cos 6\theta +1=0$ On simplifying, we get: $\Rightarrow 2+4\cos 6\theta =0$ On dividing the expression by $2$, we get: $\Rightarrow 1+2\cos 6\theta =0$ On transferring the term $1$ from the left-hand side to the right-hand side, we get: $\Rightarrow 2\cos 6\theta =-1$ On transferring the term $2$ from the left-hand side to the right-hand side, we get: $\Rightarrow \cos 6\theta =\dfrac{-1}{2}$ Now we know that $\cos x=\dfrac{-1}{2}$ when the value of $x=\dfrac{2\pi }{3}$. Now since $\cos 6\theta =\dfrac{-1}{2}$, we can conclude that: $\Rightarrow 6\theta =\dfrac{2\pi }{3}$ On transferring the term $6$ from the left-hand side to the right-hand side, we get: $\Rightarrow \theta =\dfrac{2\pi }{3\times 6}$ On simplifying, we get: $\Rightarrow \theta =\dfrac{2\pi }{18}$ On simplifying, we get: $\Rightarrow \theta =\dfrac{\pi }{9}$, which lies in the domain $\theta \in \left( 0,\pi /3 \right)$, therefore, it is the required solution. **Therefore, the correct option is $\left( C \right)$.** **Note:** It is to be remembered that in this question we have used row transformation. There also exists column transformations which use columns instead of rows. It is to be noted that doing row or column transformations on a determinant do not change its final value. The mod of the determinant is the value that the determinant holds. For a determinant $A$, the notation of its mod will be $\left| A \right|$.