Question

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

Answer 1

Hint: Try to simplify the left-hand side of the equation given in the question by the application of the sine rule of a triangle followed by the use of the formula of (sinA+sinB) and the formula of (sinA-sinB).

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.
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:
$\begin{align}
  & a=k\sin A \\
 & b=k\sin B \\
 & c=k\sin C \\
\end{align}$
So, applying this to our expression, we get
$\dfrac{b-c}{b+c}$
$=\dfrac{k\sin B-k\sin C}{k\sin B+k\sin C}$
Now we will take $k$ common from all the terms. On doing so, we get
$\dfrac{\left( \sin B-\sin C \right)}{\left( \sin B+\sin C \right)}$
Now we know that $\operatorname{sinB}-\operatorname{sinC}=2\cos \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)$ . On using this in our expression, we get
$\dfrac{2\cos \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)}{\left( \sin B+\sin C \right)}$
We also know that $\operatorname{sinB}+\operatorname{sinC}=2\sin \left( \dfrac{C+B}{2} \right)\cos \left( \dfrac{B-C}{2} \right)$ . On using this in our expression, we get
$\dfrac{\cos \left( \dfrac{C+B}{2} \right)\sin \left( \dfrac{B-C}{2} \right)}{\sin \left( \dfrac{C+B}{2} \right)\cos \left( \dfrac{B-C}{2} \right)}$
Now to reach the answer, let’s arrange the above expression according to our need.
$\dfrac{\cos \left( \dfrac{C+B}{2} \right)}{\sin \left( \dfrac{C+B}{2} \right)}\times \dfrac{sin\left( \dfrac{B-C}{2} \right)}{\cos \left( \dfrac{B-C}{2} \right)}$
$=\dfrac{\cos \left( \dfrac{C+B}{2} \right)}{\sin \left( \dfrac{C+B}{2} \right)}\times \dfrac{1}{\dfrac{\cos \left( \dfrac{B-C}{2} \right)}{\sin \left( \dfrac{B-C}{2} \right)}}$
Now we know that $\dfrac{\sin X}{\cos X}=\tan X$ . So, we can write our expression as
 $tan\left( \dfrac{C+B}{2} \right)\times \dfrac{1}{\tan \left( \dfrac{B-C}{2} \right)}$
$=\dfrac{\tan \left( \dfrac{B+C}{2} \right)}{\tan \left( \dfrac{B-C}{2} \right)}$
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. Also, you need to learn the sine rule and the cosine rule as they are used very often. The k in the sine rule is twice the radius of the circumcircle of the triangle, i.e., sine rule can also be written as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta }$ , where $\Delta $ represents the area of the triangle. Alternately, you can solve the above question using Napier’s analogy, according to which $\tan \dfrac{B-C}{2}$ is equal to $\dfrac{b-c}{b+c}\cot \dfrac{A}{2}$.