Question

In ∆ABC, prove that:
(i) $a\left(\cos B+\cos C-1\right)+b\left(\cos C+\cos A-1\right)+c\left(\cos A+\cos B-1\right)=0$

(ii) $\frac{\cos A}{b\cos C+c\cos B}+\frac{\cos B}{c\cos A+a\cos C}+\frac{\cos C}{a\cos B+b\cos A}=\frac{a^2+b^2+c^2}{2abc}$

Answer 1

Let ABC be any triangle.

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Consider}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{LHS}=a\left(\cos B+\cos C-1\right)+b\left(\cos C+\cos A-1\right)+c\left(\cos A+\cos B-1\right) \\ =a\cos B+b\cos C+a\cos C+b\cos A+c\cos A+c\cos B-\left(a+b+c\right) \\ =\left(a\cos B+b\cos A\right)+\left(b\cos C+c\cos B\right)+\left(a\cos C+c\cos A\right)-\left(a+b+c\right) \\ =c+a+b-\left(a+b+c\right)..\left(\mathrm{Using}\mathrm{projection}\mathrm{formula}:a=b\cos C+c\cos B,b=a\cos C+c\cos A,c=a\cos B+b\cos A\right) \\ =0=\mathrm{RHS} \\ \left(\mathrm{ii}\right)\mathrm{Consider}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{LHS}=\frac{\cos A}{b\cos C+c\cos B}+\frac{\cos B}{c\cos A+a\cos C}+\frac{\cos C}{a\cos B+b\cos A} \\ =\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} \\ =\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc} \\ =\frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc} \\ =\frac{a^2+b^2+c^2}{2abc}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$