Question

The value of the determinant $\left|\begin{array}{ccc}1& \cos (\alpha -\beta )& \cos \alpha \\ \cos (\alpha -\beta )& 1& \cos \beta \\ \cos \alpha & \cos \beta & 1\end{array}\right|$ is

• A. ${\alpha }^2+{\beta }^2$
• B. ${\alpha }^2-{\beta }^2$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

$0$

On solving the determinant, we have
$1(1-{\cos }^2\beta )-\cos (\alpha -\beta )\left[\cos \right(\alpha -\beta )-\cos \alpha \cdot \cos \beta ]+\cos \alpha [\cos \beta \cdot \cos (\alpha -\beta )-\cos \alpha ]$

$=1-{\cos }^2\beta -{\cos }^2\alpha -{\cos }^2(\alpha -\beta )+2\cos \alpha \cdot \cos \beta .\cos (\alpha -\beta )$

$=1-{\cos }^2\beta -{\cos }^2\alpha +\cos (\alpha -\beta )[2\cos \alpha .\cos \beta -\cos (\alpha -\beta \left)\right]$

$=1-{\cos }^2\beta -{\cos }^2\alpha +\cos (\alpha -\beta )\left[\cos \right(\alpha +\beta )+\cos (\alpha -\beta )-\cos (\alpha -\beta \left)\right]$

$=1-{\cos }^2\beta -{\cos }^2\alpha -{\cos }^2\alpha .{\cos }^2\beta -{\sin }^2\alpha .{\sin }^2\beta$

$=1-{\cos }^2\beta -{\cos }^2\alpha (1-{\cos }^2\beta )-{\sin }^2\alpha .{\sin }^2\beta$

$=1-{\cos }^2\beta -{\cos }^2\alpha .{\sin }^2\beta -{\sin }^2\alpha .{\sin }^2\beta$

$=(1-{\cos }^2\beta )-{\sin }^2\beta ({\sin }^2\alpha +{\cos }^2\alpha )$
$={\sin }^2\beta -{\sin }^2\beta =0$

Hence, determinant is $0$.