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

In any ΔABC\Delta ABC , prove that
b−cb+c=tan⁡(B−C2)tan⁡(B+C2)\dfrac{b-c}{b+c}=\dfrac{\tan \left( \dfrac{B-C}{2} \right)}{\tan \left( \dfrac{B+C}{2} \right)}

Explanation

Solution

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: 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:
a=ksin⁡A b=ksin⁡B c=ksin⁡C \begin{aligned} & a=k\sin A \\\ & b=k\sin B \\\ & c=k\sin C \\\ \end{aligned}
So, applying this to our expression, we get
b−cb+c\dfrac{b-c}{b+c}
=ksin⁡B−ksin⁡Cksin⁡B+ksin⁡C=\dfrac{k\sin B-k\sin C}{k\sin B+k\sin C}
Now we will take kk common from all the terms. On doing so, we get
(sin⁡B−sin⁡C)(sin⁡B+sin⁡C)\dfrac{\left( \sin B-\sin C \right)}{\left( \sin B+\sin C \right)}
Now we know that sinB⁡−sinC⁡=2cos⁡(C+B2)sin⁡(B−C2)\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
2cos⁡(C+B2)sin⁡(B−C2)(sin⁡B+sin⁡C)\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 sinB⁡+sinC⁡=2sin⁡(C+B2)cos⁡(B−C2)\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
cos⁡(C+B2)sin⁡(B−C2)sin⁡(C+B2)cos⁡(B−C2)\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.
cos⁡(C+B2)sin⁡(C+B2)×sin(B−C2)cos⁡(B−C2)\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)}
=cos⁡(C+B2)sin⁡(C+B2)×1cos⁡(B−C2)sin⁡(B−C2)=\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 sin⁡Xcos⁡X=tan⁡X\dfrac{\sin X}{\cos X}=\tan X . So, we can write our expression as
tan(C+B2)×1tan⁡(B−C2)tan\left( \dfrac{C+B}{2} \right)\times \dfrac{1}{\tan \left( \dfrac{B-C}{2} \right)}
=tan⁡(B+C2)tan⁡(B−C2)=\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 asin⁡A=bsin⁡B=csin⁡C=k=2R=abc2Δ\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k=2R=\dfrac{abc}{2\Delta } , where Δ\Delta represents the area of the triangle. Alternately, you can solve the above question using Napier’s analogy, according to which tan⁡B−C2\tan \dfrac{B-C}{2} is equal to b−cb+ccot⁡A2\dfrac{b-c}{b+c}\cot \dfrac{A}{2}.