Question

It is given that \[{{\Delta }_{1}}=\left| \begin{matrix}
   x & \sin \theta & \cos \theta \\
   -\sin \theta & -x & 1 \\
   \cos \theta & 1 & x \\
\end{matrix} \right|\] and \[{{\Delta }_{2}}=\left| \begin{matrix}
   x & \sin 2\theta & \cos 2\theta \\
   -\sin 2\theta & -x & 1 \\
   \cos 2\theta & 1 & x \\
\end{matrix} \right|,x\ne 0;\] then for all $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ :
$\begin{align}
  & \text{A}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=x\left( \cos 2\theta -\cos 4\theta \right) \\
 & \text{B}\text{. }{{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}} \\
 & \text{C}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=-2{{x}^{3}} \\
 & \text{D}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=-2\left( {{x}^{3}}+x-1 \right) \\
\end{align}$

Answer 1

Hint: First we have to solve the determinants given in the question. By simplifying the given determinants we find the values of ${{\Delta }_{1}}\ and {{\Delta }_{2}}$. Then, as given in the options we calculate the value of ${{\Delta }_{1}}+{{\Delta }_{2}}\And {{\Delta }_{1}}-{{\Delta }_{2}}$ and compare the values obtained with the given options to find the correct option.

Complete step-by-step answer:
We have been given \[{{\Delta }_{1}}=\left| \begin{matrix}
   x & \sin \theta & \cos \theta \\
   -\sin \theta & -x & 1 \\
   \cos \theta & 1 & x \\
\end{matrix} \right|\] and \[{{\Delta }_{2}}=\left| \begin{matrix}
   x & \sin 2\theta & \cos 2\theta \\
   -\sin 2\theta & -x & 1 \\
   \cos 2\theta & 1 & x \\
\end{matrix} \right|,x\ne 0;\]
Now, let us first simplify the given determinants to get the values of ${{\Delta }_{1}}\And {{\Delta }_{2}}$.
Now, we know that if we have determinant of order $3\times 3$ of the form \[\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\], then the value of determinant is defined as \[{{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{1}}{{c}_{3}}-{{c}_{1}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)\] .
Now, let us first consider \[{{\Delta }_{1}}=\left| \begin{matrix}
   x & \sin \theta & \cos \theta \\
   -\sin \theta & -x & 1 \\
   \cos \theta & 1 & x \\
\end{matrix} \right|\]
When we compare the given determinant with the general form, the value is defined as \[{{\Delta }_{1}}=x\left( -x\times x-1\times 1 \right)-\sin \theta \left( -\sin \theta \times x-\cos \theta \times 1 \right)+\cos \theta \left( -\sin \theta \times 1-\left( -x\times \cos \theta \right) \right)\]
Now, simplifying further we get
\[{{\Delta }_{1}}=x\left( -{{x}^{2}}-1 \right)-\sin \theta \left( -\sin \theta \times x-\cos \theta \right)+\cos \theta \left( -\sin \theta -\left( -x\times \cos \theta \right) \right)\]
\[{{\Delta }_{1}}=x\left( -{{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta \right)+\cos \theta \left( -\sin \theta +x\cos \theta \right)\]
\[{{\Delta }_{1}}=-{{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta \]
Now, simply solve the operations, we get
\[{{\Delta }_{1}}=-{{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\]
Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
So, substituting the value we get
\[\begin{align}
  & {{\Delta }_{1}}=-{{x}^{3}}-x+x\times 1 \\
 & {{\Delta }_{1}}=-{{x}^{3}}.............(i) \\
\end{align}\]
Now, let us consider second determinant \[{{\Delta }_{2}}=\left| \begin{matrix}
   x & \sin 2\theta & \cos 2\theta \\
   -\sin 2\theta & -x & 1 \\
   \cos 2\theta & 1 & x \\
\end{matrix} \right|\]
When we compare the given determinant with the general form, the value is defined as \[{{\Delta }_{2}}=x\left( -x\times x-1\times 1 \right)-\sin 2\theta \left( -\sin 2\theta \times x-\cos 2\theta \times 1 \right)+\cos 2\theta \left( -\sin 2\theta \times 1-\left( -x\times \cos 2\theta \right) \right)\]
Now, simplifying further we get
\[{{\Delta }_{2}}=x\left( -{{x}^{2}}-1 \right)-\sin 2\theta \left( -x\sin 2\theta -\cos 2\theta \right)+\cos 2\theta \left( -\sin 2\theta +x\cos 2\theta \right)\]
\[{{\Delta }_{2}}=-{{x}^{3}}-x+x{{\sin }^{2}}2\theta +\sin 2\theta \cos 2\theta -\sin 2\theta \cos 2\theta +x{{\cos }^{2}}2\theta \]
Now, simply solve the operations, we get
\[\begin{align}
  & {{\Delta }_{2}}=-{{x}^{3}}-x+x{{\sin }^{2}}2\theta +x{{\cos }^{2}}2\theta \\
 & {{\Delta }_{2}}=-{{x}^{3}}-x+x\left( {{\sin }^{2}}2\theta +{{\cos }^{2}}2\theta \right) \\
\end{align}\]
Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
So, substituting the value we get
\[\begin{align}
  & {{\Delta }_{2}}=-{{x}^{3}}-x+x\times 1 \\
 & {{\Delta }_{2}}=-{{x}^{3}}.............(ii) \\
\end{align}\]
When we add equation (i) and equation (ii), we get
\[\begin{align}
  & {{\Delta }_{1}}+{{\Delta }_{2}}=-{{x}^{3}}+\left( -{{x}^{3}} \right) \\
 & {{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}} \\
\end{align}\]
When we compare the obtained values with the options, we get that the option B is the correct answer.

Note: We added the equation (i) and equation (ii) first and we get the answer. If we didn’t get any match with the options, we subtract both the equations.
\[\begin{align}
  & {{\Delta }_{1}}-{{\Delta }_{2}}=-{{x}^{3}}-\left( -{{x}^{3}} \right) \\
 & {{\Delta }_{1}}-{{\Delta }_{2}}=-{{x}^{3}}+{{x}^{3}} \\
 & {{\Delta }_{1}}-{{\Delta }_{2}}=0 \\
\end{align}\]