Question

In any triangle $ABC$, prove that
$\frac{b^2-c^2}{a^2}\sin 2A+\frac{c^2-a^2}{b^2}\sin 2B+\frac{a^2-b^2}{c^2}\sin 2C=0$

Answer 1

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

$⟹a=k\sin A,b=k\sin B,c=k\sin C$

Now,

$LHS=\frac{(b^2-c^2)}{a^2}\sin 2A+\frac{(c^2-a^2)}{b^2}\sin 2B+\frac{(a^2-b^2)}{c^2}\sin 2C$

$LHS=\frac{(b^2-c^2)}{a^2}2\sin A\cos A+\frac{(c^2-a^2)}{b^2}2\sin B\cos B+\frac{(a^2-b^2)}{c^2}2\sin C\cos C$

$LHS=\frac{(b^2-c^2)}{a^2}\left(\frac{2a}{k}\right)\left(\frac{b^2+c^2-a^2}{2bc}\right)+\frac{(c^2-a^2)}{b^2}\left(\frac{2b}{k}\right)\left(\frac{a^2+c^2-b^2}{2ca}\right)+\frac{(a^2-b^2)}{c^2}\left(\frac{2c}{k}\right)\left(\frac{b^2-c^2+a^2}{2ba}\right)$

$LHS=\frac{1}{kabc}\left[\right(b^2-c^2\left)\right(b^2+c^2-a^2)+(c^2-a^2\left)\right(c^2+a^2-b^2)+(a^2-b^2\left)\right(a^2+b^2-c^2\left)\right]$

$LHS=\frac{1}{kabc}\left[0\right]$

$LHS=0$

$LHS=RHS$