Question

In any triangle ABC, prove the following:

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

Answer 1

Given: ABC is a triangle.

According to sine rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k...\left(i\right)$

We have to prove:

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

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

Putting the values from equation (i)

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

$=k^2\sin A\sin (B-C)+k^2\sin B\sin (C-A)+k^2\sin C\sin (A-B)$

$=k^2\left[\sin A\right(\sin B\cos C-\sin C\cos B)+\sin B(\sin C\cos A-\sin A\cos C)+\sin C(\sin A\cos B-\sin B\cos A\left)\right]$

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

= 0

= R.H.S

Hence, proved.