Question

$\frac{b^2-c^2}{a^2}sin2A+\frac{c^2-a^2}{b^2}sin2B+\frac{a^2-b^2}{c^2}sin2C=0$

Answer 1

$\left(\frac{b^2-c^2}{a^2}\right)\sin 2A+(\frac{c^2-a^2}{b^2}\sin 2B+\left(\frac{a^2-b^2}{c^2}\right))\sin 2C=0$

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

On solving $LHS$ we get

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

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

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

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