Question

In $\Delta ABC,\frac{a^2+b^2}{a^2-b^2}=\frac{\sin \left(A+B\right)}{\sin \left(A-B\right)}$, prove that the triangle is isosceles or right triangle.

Answer 1

Consider the given equation,

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

Apply the componendo and devidendo rule .

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

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

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

$\because \sin c+\sin d=2\sin \frac{c+d}{2}\cos \frac{c-d}{2},\sin c-\sin d=2\cos \frac{c+d}{2}\sin \frac{c-d}{2}$

$\therefore \frac{a^2}{b^2}=\frac{2\sin A\cos B}{2\cos A\sin B}$

By $\sin$ rule, $a=R\sin A,b=R\sin B$

$\frac{{\left(R\sin A\right)}^2}{{\left(R\sin B\right)}^2}=\frac{2\sin A\cos B}{2\cos A\sin B}$

$\frac{\sin A}{\sin B}=\frac{\cos B}{\cos A}$

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

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

$\sin 2A-\sin 2B=0$

$2\cos \left(\frac{2A+2B}{2}\right)\sin \left(\frac{2A-2B}{2}\right)=0$

$2\cos (A+B)\sin (A-B)=0$

$\cos (A+B)\sin (A-B)=0$

If $\cos (A+B)=0$

$\sin (A+B)=\cos {90}^0$

$A+B={90}^0$

Hence, triangle is right angle triangle.

If, $\sin A-B=0$

$\sin A-B=\sin 0$

$A-B=0$

$A=B$

So, the triangle is isosceles.

Hence, proved.