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In $\triangle ABC$

$ \dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)} $

then what type of triangle is $\triangle ABC $ ?

My try :

By componendo and dividendo

$\dfrac{a^2}{b^2} = \tan A \cot B$

Not able to conclude, any help ?

AgentS
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2 Answers2

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You are nearly there. Just apply the sine rule in the form $$\dfrac ab=\dfrac{\sin A}{\sin B}$$to get $\sin A\cos A=\sin B\cos B$; that is, $\sin 2A=\sin 2B$ and hence either $A=B$ or $C$ ($=\pi-A-B$) =$\pi/2$.

Edit Apologies to the OP for an incorrect first answer.

John Bentin
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  • Wow! how did I miss that !! thank you so much xD – AgentS Aug 22 '14 at 09:18
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    I am definately sure you reached to this point @John Bentin, what after that.$$\dfrac{a}{b} = \frac{\cos B}{\cos A}$$ – MonK Aug 22 '14 at 09:49
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    @MonK: You are right. See my corrected answer. – John Bentin Aug 22 '14 at 09:52
  • Can you elaborate your answer a little more, I am sorry. – MonK Aug 22 '14 at 09:54
  • @MonK: I have done so. Hope it's clear now. – John Bentin Aug 22 '14 at 10:02
  • It still doesn't, Look below$$a\cos A = b\cos B$$

    $$a\cos A = b\cos (180-(A+C))$$

    $$a\cos A = b(-\cos(A+C))$$

    $$a\cos A=-b(\cos A\cos C-\sin A\sin C)$$

    ??

    I am pretty sure I went in the wrong direction after it, Can you guide me please.

    How did you reach to $sinAcosA=sinBcosB$

     – MonK Aug 22 '14 at 10:14
  • One more thing, We can be definately sure that $A+B+C=180^\circ$, but $A+B=90^\circ$, It can't be established. Don't take this personal, I am trying to understand your answer :) – MonK Aug 22 '14 at 10:16
  • @MonK: In your first line, you need to replace $a$ and $b$ by $\sin A$ and $\sin B$ respectively, which you can do by the sine rule. From $\sin 2A=\sin 2B$ we get either $2A=2B$ or $2A +2B=\pi$ (because $\sin(\pi-\theta)=\sin\theta$). – John Bentin Aug 22 '14 at 10:40
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$$\dfrac{a^2-b^2}{a^2+b^2} = \dfrac{\sin(A-B)}{\sin(A+B)}$$ $$\dfrac{a^2-b^2}{a^2+b^2} = \frac{\sin A\cos B-\cos A\sin B}{\sin A\cos B+\cos A\sin B}$$

$$\dfrac{a^2-b^2+a^2+b^2}{a^2-b^2-a^2-b^2} = \frac{\sin A\cos B-\cos A\sin B+\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B-\sin A\cos B-\cos A\sin B}$$

$$\dfrac{a^2}{b^2} = \frac{\sin A\cos B}{\cos A\sin B}$$ $$\dfrac{a^2}{b^2} = \frac{ak.\cos B}{\cos A.bk}$$

Because,

$$\text{sine rule}\implies\frac{\sin A}a=\frac{\sin B}b=k$$

$$\dfrac{a}{b} = \frac{\cos B}{\cos A}$$

$$a\cos A = b\cos B$$

$$\sin A\cos A=\sin B\cos B$$

$$\sin 2A=\sin 2B$$

Or, A=B.

So, the triangle is an isoceles.

MonK
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