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# 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
Trigonometry

7 out of 10

##### Suggested Book
Plane Trigonometry by S.L. Loney

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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