Question

If $\alpha ,\beta$ and $\gamma$ are real numbers, the $A=\left|\begin{array}{ccc}1& \cos (\beta -\alpha )& \cos (\gamma -\alpha )\\ \cos (\alpha -\beta )& 1& \cos (\gamma -\beta )\\ \cos (\alpha -\gamma )& \cos (\beta -\gamma )& 1\end{array}\right|$ is equal to.

• A. $1$
• B. $-1$
• C. $\cos \alpha \cos \beta \cos \gamma$
• D. $0$

Answer 1

The correct option is D $0$

Let $P=\beta -\alpha$

$\therefore \cos (\beta -\alpha )=\cos (\alpha -\beta )=\cos P$

$Q=\alpha -r$

& $R=r-\beta$

$\therefore A=\left|\begin{array}{ccc}1& \cos P& \cos Q\\ \cos P& 1& \cos R\\ \cos Q& \cos R& 1\end{array}\right|$

where $P+Q+R=0$

$(1-{\cos }^{2}R)-\cos P[\cos P-\cos Q\cos R]+\cos Q[\cos P\cos R-\cos Q]$

$\Rightarrow 1+2\cos Q\cos R\cos P-[{\cos }^{2}P+{\cos }^{2}Q+{\cos }^{2}R]$

As ${\cos }^{2}P+{\cos }^{2}Q+{\cos }^{2}R=1+2\cos Q\cos R\cos P$

We get, $A=0$