Understand the problem

Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. If \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\), then prove that the triangle is isosceles.

Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016 
Topic
Trigonometry

Difficulty Level
7 out of 10

Suggested Book
 Plane Trigonometry by S.L. Loney

Start with hints

Do you really need a hint? Try it first!

Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. Given \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\)    

\(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})\) \(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\)  

\(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\) \(\Rightarrow  b\sin A\cos B=a\cos A \sin B\) \(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \)

 

\(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}\)  [since \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)] \(\Rightarrow \frac{\cos B}{\cos A}=1\) \(\cos B=\cos A\) \(A=B\).

\(\Rightarrow \Delta ABC\) is isosceles  (Proved).

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