Question Number 90535 by 1223 last updated on 24/Apr/20
$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC},\:{a}\left({b}\:\mathrm{cos}\:{C}−{c}\:\mathrm{cos}\:{B}\right)= \\ $$
Answered by $@ty@m123 last updated on 24/Apr/20
$${a}\left({b}×\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}−{c}×\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\right) \\ $$$${a}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{a}}−\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{a}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{c}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} \right) \\ $$$${b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$