Question

In any $\Delta ABC$, prove that ${{a}^{2}}\sin \left( B-C \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\sin A$.

Answer 1

Hint: We will be using the concept of solution of triangles to solve the problem. We will use the sine rule to relate the sides a and c with the sine of angle A and C and similarly we will find the relation between the sides a and b then we will use these equations in the left hand side of the equation.

Complete Step-by-Step solution:
We have been given a$\Delta ABC$ and we have to prove that, ${{a}^{2}}\sin \left( B-C \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\sin A$.
So, we will first draw a $\Delta ABC$ and label its sides as a, b, c.

Now, we will take L.H.S of the question and prove it to be equal to the R.H.S.
In L.H.S we have,
${{a}^{2}}\sin \left( B-C \right)$
Now, we know that the trigonometric identity that,
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin A$
So, we have,
${{a}^{2}}\sin \left( B-C \right)={{a}^{2}}\sin B\cos C-{{a}^{2}}\cos B\sin C$
Now, we know that according to sine rule,
$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k$
So, we have,
$\begin{align}
  & a\sin C=c\sin A..........\left( 1 \right) \\
 & a\sin B=b\sin A..........\left( 2 \right) \\
\end{align}$
Now, we will use (1) and (2) in,
${{a}^{2}}\sin B\cos C-{{a}^{2}}\cos B\sin C$
So, we have,
$\begin{align}
  & {{a}^{2}}\sin \left( B-C \right)=ab\sin A\cos C-ac\cos B\sin A \\
 & =a\sin A\left( ab\cos C-ac\cos B \right) \\
\end{align}$
Now, we know that according to cosine rule we have,
$\begin{align}
  & \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}.........\left( 3 \right) \\
 & \cos B=\dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}.........\left( 4 \right) \\
\end{align}$
Now, using (3) and (4) above we have,
$\begin{align}
  & {{a}^{2}}\sin \left( B-C \right)=\sin A\left( ab\dfrac{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2ab}-ac\dfrac{\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2ac} \right) \\
 & =\sin A\left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2}-\dfrac{\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2} \right) \\
 & =\sin A\left( \dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{c}^{2}}-{{a}^{2}}+{{b}^{2}}}{2} \right) \\
\end{align}$
Now, we will further simplify the numerator,
$\begin{align}
  & =\sin A\left( \dfrac{2{{b}^{2}}-2{{c}^{2}}}{2} \right) \\
 & =\sin A\left( {{b}^{2}}-{{c}^{2}} \right) \\
 & {{a}^{2}}\sin \left( B-C \right)=\left( {{b}^{2}}-{{c}^{2}} \right)\sin A \\
\end{align}$
Since L.H.S = R.H.S, Hence proved.

Note: To solve these types of questions it is important to note that we have used the projection formulae of the triangle to find the relation in equation (1) and (2) as follows.
$\begin{align}
  & a\sin C=c\sin A..........\left( 1 \right) \\
 & a\sin B=b\sin A..........\left( 2 \right) \\
\end{align}$
also it is important to note that we have used the trigonometric identity that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin A$ to further simplify the left hand side of the equation.