In a ∆ A B C , 2 ( b c cos A + c a cos B + a b cos C ) = ?

Given that, the trigonometric expression is $2(bc\cos A+ca\cos B+ab\cos C)$

$=2bc\cos A+2ca\cos B+2ab\cos C$

We know that, $\cos A=\frac{\left(b^2+c^2-a^2\right)}{2bc}$.

$\Rightarrow 2bc·\frac{\left(b^2+c^2-a^2\right)}{2bc}+2ca·\frac{\left(c^2+a^2-b^2\right)}{2ca}+2ab·\frac{\left(a^2+b^2-c^2\right)}{2ab}$

$=\left(b^2+c^2-a^2\right)+\left(c^2+a^2-b^2\right)+\left(a^2+b^2-c^2\right)$

$=\left(a^2+b^2+c^2\right)$

Hence, the value of the given trigonometric expression is $\left(a^2+b^2+c^2\right)$.