Question

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

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 values of cosC and cosB with the sides of the triangle then we will substitute the value in the left hand side of the equation to convert the whole left hand side in terms of the sides of the triangle and further simplify it to prove the given problem.

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

Now, we know that in $\Delta ABC$ according to cosine rule we have,
$\begin{align}
  & \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}..........\left( 1 \right) \\
 & \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}..........\left( 2 \right) \\
\end{align}$
Now, we will take the left – hand side of the equation we have to prove and prove it equal to the right hand side. So, we have,
$a\left( b\cos C-c\cos B \right)$
Now, we will substitute equation (1) & (2) in it,
$\begin{align}
  & a\left( b\cos C-c\cos B \right)=a\left( b\dfrac{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2ab}-c\dfrac{\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}{2ac} \right) \\
 & =ab\dfrac{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2ab}-ac\dfrac{\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}{2ac} \\
\end{align}$
Now, we will solve this further we have,
$\begin{align}
  & a\left( b\cos C-c\cos B \right)=\dfrac{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2}-\dfrac{\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}{2} \\
 & =\dfrac{1}{2}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{a}^{2}}-{{c}^{2}}+{{b}^{2}} \right) \\
 & =\dfrac{1}{2}\left( 2{{b}^{2}}-2{{c}^{2}} \right) \\
\end{align}$
Now, we will cancel 2 in both numerator and denominator,
$a\left( b\cos C-c\cos B \right)={{b}^{2}}-{{c}^{2}}$
Since, we have L.H.S = R.H.S
Hence Proved.

Note: To solve these types of questions it is important to note that we have used cosine rule to relate the cosine of angles with the sides of the triangles also it is important to note that we have done so because in the right side of the equation to be proved we have only sides of the triangle.