Question

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

Answer 1

Hint: Try to simplify the left-hand side of the equation given in the question using the formula of sin(X-Y) followed by the application of the sine rule and the cosine rule of a triangle.

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now starting with the left-hand side of the equation that is given in the question. To start the simplification, we need to use the formula sin(X-Y) = cosYsinX – sinYcosX.
$\dfrac{\sin \left( B-C \right)}{\sin \left( B+C \right)}$
\[=\dfrac{\sin B\cos C-\cos B\sin C}{\sin B\cos C+\cos B\sin C}\]
Now we know, according to the sine rule of the triangle: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$ and in other terms, it can be written as:
$\begin{align}
  & \dfrac{a}{k}=\sin A \\
 & \dfrac{b}{k}=\sin B \\
 & \dfrac{c}{k}=\sin C \\
\end{align}$
So, applying this to our expression, we get
\[=\dfrac{\dfrac{b}{k}\cos C-\dfrac{c}{k}\cos B}{\dfrac{b}{k}\cos C+\dfrac{c}{k}\cos B}\]
Now we will take $\dfrac{1}{k}$ common from all the terms. On doing so, we get
\[=\dfrac{\dfrac{1}{k}\left( b\cos C-c\cos B \right)}{\dfrac{1}{k}\left( b\cos C+c\cos B \right)}\]
Now from the conditions of sine rule, we know that k is a finite number, so we can say that $\dfrac{1}{k}$ is finite. Therefore, cancelling $\dfrac{1}{k}$ from denominator and numerator, our expression becomes:
 \[=\dfrac{b\cos C-c\cos B}{b\cos C+c\cos B}\]
Now according to the cosine rule of a triangle:
$\begin{align}
  & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \\
 & \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} \\
 & \cos C=\dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2ab} \\
\end{align}$
So, using this in our expression, we get
\[=\dfrac{\left( \dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2a} \right)-\left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2a} \right)}{\left( \dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2a} \right)+\left( \dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2a} \right)}\]
Now we will take $\dfrac{1}{2a}$ common from all the terms and cancel it. On doing so, we get
\[=\dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}-{{a}^{2}}-{{c}^{2}}+{{b}^{2}}}{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}+{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}\]
\[=\dfrac{2{{b}^{2}}-2{{c}^{2}}}{2{{a}^{2}}}\]
\[=\dfrac{{{b}^{2}}-{{c}^{2}}}{{{a}^{2}}}\]
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. A general mistake that students make is completely wrong. Also, you need to learn the sine rule and cosine rule, as they are often used.