Question

Find the value of the determinant
\[\left| \begin{matrix}
   \cos \left( x-y \right) & \cos \left( y-z \right) & \cos \left( z-x \right) \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
\[\begin{align}
  & A.2\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( z-x \right) \\
 & B.-2\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( z-x \right) \\
 & C.2\cos \left( x-y \right)\cos \left( y-z \right)\cos \left( z-x \right) \\
 & D.-2\cos \left( x-y \right)\cos \left( y-z \right)\cos \left( z-x \right) \\
\end{align}\]

Answer 1

Hint: In this question, we have to find a determinant whose entries are of sine and cosine function. For this, we will first simplify the determinant using various operations and then find the determinant. We will use various trigonometric properties in this question which are:
\[\begin{align}
  & \left( i \right)\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right) \\
 & \left( ii \right)\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\
 & \left( iii \right)\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y \\
 & \left( iv \right)\cot x=\dfrac{\sin x}{\cos y} \\
 & \left( v \right)\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y \\
 & \left( vi \right)\sin \left( -x \right)=-\sin x \\
\end{align}\]

Complete step by step answer:
Here we are given the determinant as,
\[\left| \begin{matrix}
   \cos \left( x-y \right) & \cos \left( y-z \right) & \cos \left( z-x \right) \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Before directly finding the values of determinant, let us first simplify it using some properties.
First of all, let us use the operation on row first that ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ we get:
\[\left| \begin{matrix}
   \cos \left( x-y \right)-\cos \left( x+y \right) & \cos \left( y-z \right)-\cos \left( y+z \right) & \cos \left( z-x \right)-\cos \left( z+x \right) \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Now we know that $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ so let us apply it on first row, we get:
\[\left| \begin{matrix}
   2\sin \left( \dfrac{x-y+x+y}{2} \right)\sin \left( \dfrac{x+y-x+y}{2} \right) & 2\sin \left( \dfrac{y-z+y+z}{2} \right)\sin \left( \dfrac{y+z-y+z}{2} \right) & 2\sin \left( \dfrac{z-x+z+x}{2} \right)\sin \left( \dfrac{z+x-z+x}{2} \right) \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Simplifying we get:
\[\left| \begin{matrix}
   2\sin x\sin y & 2\sin y\sin z & 2\sin x\sin z \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Taking 2 common from the first row (we can take any common factor out of any row or column in a determinant matrix), we get:
\[2\left| \begin{matrix}
   \sin x\sin y & \sin y\sin z & \sin x\sin z \\
   \cos \left( x+y \right) & \cos \left( y+z \right) & \cos \left( z+x \right) \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
We know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ so let us apply to the second row of determinant, we get:
\[2\left| \begin{matrix}
   \sin x\sin y & \sin y\sin z & \sin x\sin z \\
   \cos x\cos y-\sin x\sin y & \cos y\cos z-\sin y\sin z & \cos z\cos x-\sin x\sin z \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Now let us use row operation on second row which is ${{R}_{2}}\to {{R}_{1}}+{{R}_{2}}$ we get:
\[2\left| \begin{matrix}
   \sin x\sin y & \sin y\sin z & \sin x\sin z \\
   \cos x\cos y & \cos y\cos z & \cos z\cos x \\
   \sin \left( x+y \right) & \sin \left( y+z \right) & \sin \left( z+x \right) \\
\end{matrix} \right|\]
Now, let us take sinxsiny common from first column, sinysinz common from second column and sinzsinx common from third column we get:
\[2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left| \begin{matrix}
   1 & 1 & 1 \\
   \dfrac{\cos x\cos y}{\sin x\sin y} & \dfrac{\cos y\cos z}{\sin y\sin z} & \dfrac{\cos z\cos x}{\sin x\sin z} \\
   \dfrac{\sin \left( x+y \right)}{\sin x\sin y} & \dfrac{\sin \left( y+z \right)}{\sin y\sin z} & \dfrac{\sin \left( z+x \right)}{\sin x\sin z} \\
\end{matrix} \right|\]
We know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ so applying it on second row we get:
\[2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left| \begin{matrix}
   1 & 1 & 1 \\
   \cot x\cot y & \cot y\cot z & \cot z\cot x \\
   \dfrac{\sin \left( x+y \right)}{\sin x\sin y} & \dfrac{\sin \left( y+z \right)}{\sin y\sin z} & \dfrac{\sin \left( z+x \right)}{\sin x\sin z} \\
\end{matrix} \right|\]
Now let us solve the general formula $\dfrac{\sin \left( A+B \right)}{\sin A\sin B}$ we know $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$ we get,
\[\begin{align}
  & \dfrac{\sin \left( A+B \right)}{\sin A\sin B}=\dfrac{\sin A\cos B+\sin B\cos A}{\sin A\sin B} \\
 & \Rightarrow \dfrac{\sin A\cos B}{\sin A\sin B}+\dfrac{\cos A\sin B}{\sin A\sin B} \\
 & \Rightarrow \dfrac{\cos B}{\sin B}+\dfrac{\cos A}{\sin A} \\
 & \Rightarrow \cot B+\cot A \\
\end{align}\]
So $\dfrac{\sin \left( A+B \right)}{\sin A\sin B}=\cot B+\cot A$ applying this on third row we get:
\[2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left| \begin{matrix}
   1 & 1 & 1 \\
   \cot x\cot y & \cot y\cot z & \cot z\cot x \\
   \cot x+\cot y & \cot y+\cot z & \cot z+\cot x \\
\end{matrix} \right|\]
Now applying column operations on column 1 and column 2 which are ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}\text{ and }{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$ we get:
\[2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left| \begin{matrix}
   0 & 0 & 1 \\
   \cot y\left( \cot x-\cot z \right) & \cot z\left( \cot y-\cot x \right) & \cot z\cot x \\
   \cot x-\cot z & \cot y-\cot x & \cot z+\cot x \\
\end{matrix} \right|\]
Taking cotx-cotz common from first column and coty-cotz from second column we get:
\[2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left( \cot x-\cot z \right)\left( \cot y-\cot x \right)\left| \begin{matrix}
   0 & 0 & 1 \\
   \cot y & \cot z & \cot z\cot x \\
   1 & 1 & \cot z+\cot x \\
\end{matrix} \right|\]
Now expanding using row 1 we get:
\[\begin{align}
  & 2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left( \cot x-\cot z \right)\left( \cot y-\cot x \right)\left[ 1\left( \cot y-\cot z \right) \right] \\
 & \Rightarrow 2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left( \cot x-\cot z \right)\left( \cot y-\cot x \right)\left( \cot y-\cot z \right) \\
\end{align}\]
We know $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ so we get:
\[\Rightarrow 2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left( \dfrac{\cos x}{\sin x}-\dfrac{\cos z}{\sin z} \right)\left( \dfrac{\cos y}{\sin y}-\dfrac{\cos x}{\sin x} \right)\left( \dfrac{\cos y}{\sin y}-\dfrac{\cos z}{\sin z} \right)\]
Taking LCM we get:
\[\Rightarrow 2{{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z\left( \dfrac{\cos x\sin z-\cos z\sin x}{\sin x\sin z} \right)\left( \dfrac{\cos y\sin x-\cos x\sin y}{\sin y\sin x} \right)\left( \dfrac{\cos y\sin z-\cos z\sin y}{\sin y\sin z} \right)\]
Cancelling ${{\sin }^{2}}x{{\sin }^{2}}y{{\sin }^{2}}z$ from the numerator and denominator we get:
\[\Rightarrow 2\left( \cos x\sin z-\cos z\sin x \right)\left( \cos y\sin x-\cos x\sin y \right)\left( \cos y\sin z-\cos z\sin y \right)\]
We know that $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$ so we get:
\[\Rightarrow 2\sin \left( z-x \right)\sin \left( x-y \right)\sin \left( z-y \right)\]
Taking negative sign common in sin(z-y) and using $\sin \left( -\theta \right)=-\sin \theta $ we get:
\[\begin{align}
  & \Rightarrow 2\sin \left( x-y \right)\sin \left( -\left( y-z \right) \right)\sin \left( z-x \right) \\
 & \Rightarrow -2\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( z-x \right) \\
\end{align}\]

So, the correct answer is “Option B”.

Note: Students should keep in mind all the trigonometric identities for solving this question. Take care of positive and negative signs while solving this sum. This sum involves a lot of complex calculations, so do it step by step. It is important to mention the operation which you are going to perform.