Question

Find the value of the determinant $\left|\begin{array}{ccc}cos(x-y)& cos(y-z)& cos(z-x)\\ cos(x+y)& cos(y+z)& cos(z+x)\\ sin(x+y)& sin(y+z)& sin(z+x)\end{array}\right|$

• A. $2\sin (x-y)\sin (y-z)\sin (x-z)$
• B. $4\sin (x-y)\sin (y-z)\sin (x-z)$
• C. $8\sin (x-y)\sin (y-z)\sin (x-z)$
• D. None of these
  

Answer 1

Answer: (A)

$2\sin (x-y)\sin (y-z)\sin (x-z)$

$\left|\begin{array}{ccc}\cos (x-y)& \cos (y-z)& \cos (z-x)\\ \cos (x+y)& \cos (y+z)& \cos (z+x)\\ \sin (x+y)& \sin (y+z)& \sin (z+x)\end{array}\right|$
$R_1\to R_1-R_2$

$=2\left|\begin{array}{ccc}\sin x\sin y& \sin y\sin z& \sin x\sin z\\ \cos (x+y)& \cos (y+x)& \cos (z+x)\\ \sin (x+y)& \sin (y+z)& \sin (x+z)\end{array}\right|$

$R_2\to R_1+R_2$

$=\left|\begin{array}{ccc}\sin x\sin y& \sin y\sin z& \sin x\sin z\\ \cos x\cos y& \cos y\cos x& \cos x\cos z\\ \sin (x+y)& \sin (y+z)& \sin (x+z)\end{array}\right|$

Taking $\sin x\sin y$ common from $C_1$, $\sin y\sin z$ from $C_2$, $\sin x\sin z$ from $C_3$.

$=2{\sin }^{2}x{\sin }^{2}y{\sin }^{2}z\left|\begin{array}{ccc}1& 1& 1\\ \cot x\cot y& \cot y\cot z& \cot z\cot x\\ \cot x+\cot y& \cot y+\cot z& \cot z+\cot x\end{array}\right|$

$[\therefore \frac{\sin (x+y)}{\sin x\sin y}=\cot x+\cot y]$
$C_1\to C_1-C_2,C_2\to C_2-C_3$

$=2{\sin }^{2}x{\sin }^{2}y{\sin }^{2}z\left|\begin{array}{ccc}0& 0& 1\\ \cot y(\cot x-\cot z)& \cot z(\cot y-\cot x)& \cot z\cot x\\ \cot x-\cot z& \cot y-\cot x& \cot z+\cot x\end{array}\right|$
Taking $(\cot x-\cot z)$ common from $C_1$ and $\cot y-\cot x$ common from $C_2$

$=2{\sin }^{2}x{\sin }^{2}y{\sin }^{2}z(\cot x-\cot z)(\cot y-\cot x)\left|\begin{array}{ccc}0& 0& 1\\ \cot y& \cot z& \cot z\cot x\\ 1& 1& \cot z+\cot x\end{array}\right|$

On expanding, we get

$=2{\sin }^{2}x{\sin }^{2}y{\sin }^{2}z(\cot x-\cot z)(\cot y-\cot x)(\cot y-\cot z)$

$=2\sin (x-y)\sin (y-z)\sin (x-z)$