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Question: In any \(\Delta ABC\), prove that \(\dfrac{\sin \left( B-C \right)}{\sin \left( B+C \right)}=\dfra...

In any ΔABC\Delta ABC, prove that
sin⁡(B−C)sin⁡(B+C)=b2−c2a2\dfrac{\sin \left( B-C \right)}{\sin \left( B+C \right)}=\dfrac{{{b}^{2}}-{{c}^{2}}}{{{a}^{2}}}

Explanation

Solution

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.
sin⁡(B−C)sin⁡(B+C)\dfrac{\sin \left( B-C \right)}{\sin \left( B+C \right)}
=sin⁡Bcos⁡C−cos⁡Bsin⁡Csin⁡Bcos⁡C+cos⁡Bsin⁡C=\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: asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k and in other terms, it can be written as:
ak=sin⁡A bk=sin⁡B ck=sin⁡C \begin{aligned} & \dfrac{a}{k}=\sin A \\\ & \dfrac{b}{k}=\sin B \\\ & \dfrac{c}{k}=\sin C \\\ \end{aligned}
So, applying this to our expression, we get
=bkcos⁡C−ckcos⁡Bbkcos⁡C+ckcos⁡B=\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 1k\dfrac{1}{k} common from all the terms. On doing so, we get
=1k(bcos⁡C−ccos⁡B)1k(bcos⁡C+ccos⁡B)=\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 1k\dfrac{1}{k} is finite. Therefore, cancelling 1k\dfrac{1}{k} from denominator and numerator, our expression becomes:
=bcos⁡C−ccos⁡Bbcos⁡C+ccos⁡B=\dfrac{b\cos C-c\cos B}{b\cos C+c\cos B}
Now according to the cosine rule of a triangle:
cos⁡A=b2+c2−a22bc cos⁡B=a2+c2−b22ac cos⁡C=b2+a2−c22ab \begin{aligned} & \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{aligned}
So, using this in our expression, we get
=(b2+a2−c22a)−(a2+c2−b22a)(b2+a2−c22a)+(a2+c2−b22a)=\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 12a\dfrac{1}{2a} common from all the terms and cancel it. On doing so, we get
=b2+a2−c2−a2−c2+b2b2+a2−c2+a2+c2−b2=\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}}}
=2b2−2c22a2=\dfrac{2{{b}^{2}}-2{{c}^{2}}}{2{{a}^{2}}}
=b2−c2a2=\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.