Question

$\frac{co{s}^{2}B-co{s}^{2}C}{b+c}+\frac{co{s}^{2}C-co{s}^{2}A}{c+a}+\frac{co{s}^{2}A-co{s}^{2}B}{a+b}=0$

Answer 1

$LHS\frac{co{s}^{2}B-co{s}^{2}C}{b+c}+\frac{co{s}^{2}C-co{s}^{2}A}{c+a}+\frac{co{s}^{2}A-co{s}^{2}B}{a+b}=\frac{co{s}^{2}B-co{s}^{2}C}{b+c}+\frac{co{s}^{2}C-co{s}^{2}A}{c+a}+\frac{co{s}^{2}A-co{s}^{2}B}{a+b}=\frac{1-si{n}^{2}B-1+si{n}^{2}C}{b+c}+\frac{1-si{n}^{2}C-1+si{n}^{2}A}{c+a}+\frac{1-si{n}^{2}A-1+si{n}^{2}B}{a+b}=\frac{si{n}^{2}C-si{n}^{2}B}{b+c}+\frac{si{n}^{2}A-si{n}^{2}C}{c+a}+\frac{si{n}^{2}B-si{n}^{2}A}{a+b}=\frac{k^2{c}^2-k^2{b}^2}{b+c}+\frac{k^2{a}^2-k^2{c}^2}{c+a}+\frac{k^2{b}^2-k^2{a}^2}{a+b}=k^2(\frac{c^2-b^2}{b+c}+\frac{a^2-c^2}{c+a}+\frac{b^2-a^2}{a+b})=k^2(c-b+a-c+b-a)[\text{Using}{b}^2-a^2=(b-a)(b+a)]=0=RHS\text{Hence proved.}$