Question

In any triangle ABC, prove the following:

$a^2\sin (B-C)=(b^2-c^2)\sin A$

Answer 1

Given: ABC is a triangle,

$[\therefore A+B+C=\pi ]$

According to sine rule,

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

We have to prove:

$a^2\sin (B-C)=(b^2-c^2)\sin A$

$R.H.S=(b^2-c^2)\sin A$

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

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

$[\because {\sin }^{2}A-{\sin }^{2}B=\sin (A+B\left)\sin \right(A-B\left)\right]$

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

$[\because A+B+C=\pi ]$

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

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

$=a^2\sin (B-C)$

$[\because a=k\sin A]$

= L.H.S

Hence, proved.