In any triangle $\mathrm{ABC}, a(b \cos C-c \cos B)=$
In any triangle $\mathrm{ABC}, a(b \cos C-c \cos B)=$
$b-c$
$b+c$
$b^2-c^2$
$b^2+c^2$
Solution
Given expression a (bcosc $-\cos B)$.
$
\begin{aligned}
& a b \cos C-a c c o s B=a b \frac{\left(a^2+b^2-c^2\right)}{2 a b}-a c \frac{\left(a^2+c^2-b^2\right)}{2 a c} \\
& =\frac{1}{2}\left[a^2+b^2-c^2-a^2-c^2+b^2\right] \\
& =\frac{1}{2}\left[2\left(b^2-c^2\right)\right]=b^2-c^2
\end{aligned}
$