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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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