Question

In $\Delta ABC,a^2({\cos }^{2}B-{\cos }^{2}C)+b^2({\cos }^{2}C-{\cos }^{2}A)+c^2({\cos }^{2}A-{\cos }^{2}B)=$

Answer 1

$a^2(co{s}^{2}B-co{s}^{2}C)+b^2(co{s}^{2}C-co{s}^{2}A)+c^2(co{s}^{2}A-co{s}^{2}B)$

$=a^{2}co{s}^{2}B-a^{2}co{s}^{2}C+b^{2}co{s}^{2}C-b^{2}co{s}^{2}A+c^{2}co{s}^{2}A-c^{2}co{s}^{2}B$

$=(a^{2}co{s}^{2}B-b^{2}co{s}^{2}A)+(b^{2}co{s}^{2}C-c^{2}co{s}^{2}B)+(c^{2}co{s}^{2}A-a^{2}co{s}^{2}C)$

$=(a\cos B+b\cos A)(a\cos B-b\cos A)+(b\cos C+c\cos B)(b\cos C-c\cos B)$
$+(c\cos A+a\cos C)(c\cos A-a\cos C)$

$=\left(c\right)(a\cos B-b\cos A)+\left(a\right)(b\cos C-c\cos B)+\left(b\right)(c\cos A-a\cos C)$

$=ac\cos B-bc\cos A+ab\cos C-ac\cos B+bc\cos A-ab\cos C$

$=0$