Question

Prove that:
$\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}=0$

Answer 1

$L.H.S.$

We know that,

Using sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\Rightarrow a=k\sin A$

$\Rightarrow b=k\sin B$

$\Rightarrow c=k\sin C$

So, putting in LHS.

$\frac{k^2{\sin }^{2}A-k^2{\sin }^{2}B}{\cos A+\cos B}+\frac{k^2{\sin }^{2}B-k^2{\sin }^{2}C}{\cos B+\cos C}+\frac{k^2{\sin }^{2}C-k^2{\sin }^{2}A}{\cos C+\cos A}$

$\Rightarrow k^2(\frac{{\sin }^{2}A-{\sin }^{2}B}{\cos A+\cos B}+\frac{{\sin }^{2}B-{\sin }^{2}C}{\cos B+\cos C}+\frac{{\sin }^{2}C-{\sin }^{2}A}{\cos C+\cos A})$

$\Rightarrow k^2(\frac{1-{\cos }^{2}A-1+{\cos }^{2}B}{\cos A+\cos B}+\frac{1-{\cos }^{2}B-1+{\cos }^{2}C}{\cos B+\cos C}+\frac{1-{\cos }^{2}C-1+{\cos }^{2}A}{\cos C+\cos A})$

$\Rightarrow k^2(\frac{{\cos }^{2}B-{\cos }^{2}A}{\cos A+\cos B}+\frac{{\cos }^{2}C-{\cos }^{2}B}{\cos B+\cos C}+\frac{{\cos }^{2}A-{\cos }^{2}C}{\cos C+\cos A})$

$\Rightarrow k^2(\frac{(\cos B-\cos A)(\cos A+\cos B)}{\cos A+\cos B}+\frac{(\cos C-\cos B)(\cos B+\cos C)}{\cos B+\cos C}+\frac{(\cos A-\cos C)(\cos C+\cos A)}{\cos C+\cos A})$

$\Rightarrow k^2(\cos B-\cos A+\cos C-\cos B+\cos A-\cos C)$

$\Rightarrow k^2\times 0=0$

$R.H.S.$

Hence, this is the answer.