Question

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

Answer 1

Hint: We will be using the concept of solution of triangles to solve the problem. We will be using cosine rules to relate the cosine of angle B and A with the sides of the triangle then we will use this value of cosB and cosA and substitute it in the left hand side of the equation to be proved then we will further simplify the equation to prove it to be equal to right hand side.

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

Now, we know that according to cosine rule,
$\begin{align}
  & \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}..........\left( 1 \right) \\
 & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}..........\left( 2 \right) \\
\end{align}$
Now, we will take the left – hand side and prove it equal to the right hand side. So, we have L.H.S as,
$ac\cos B-bc\cos A$
Now, we will substitute equation (1) & (2), so, we get,
\[\begin{align}
  & =ac\left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} \right)-bc\left( \dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \right) \\
 & ac\cos B-bc\cos A=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2}-\dfrac{\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}{2} \\
 & =\dfrac{1}{2}\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}}-{{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right) \\
 & =\dfrac{1}{2}\left( 2{{a}^{2}}-2{{b}^{2}} \right) \\
 & ={{a}^{2}}-{{b}^{2}} \\
 & ac\cos B-bc\cos A=\left( {{a}^{2}}-{{b}^{2}} \right) \\
\end{align}\]
Since, we have L.H.S = R.H.S
Hence, \[ac\cos B-bc\cos A=\left( {{a}^{2}}-{{b}^{2}} \right)\] is proved.

Note: To solve these types of questions it is important to notice that we have used cosine rule to relate the value of the cosB and cosA with the sides of the triangle because in the right hand side of the equation to be proved we have only sides of triangle therefore we have used the cosine rule.