Question

If in a $\Delta ABC,\frac{a^2-b^2}{a^2+b^2}=\frac{sin(A-B)}{s}in(A+B)$, prove that it is either a right-angled or an isosceles triangle.

Answer 1

$Accordingtosinerule,\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=k$

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

$\Rightarrow \frac{k^{2}si{n}^{2}A-k^{2}si{n}^{2}B}{k^{2}si{n}^{2}A+k^{2}si{n}^{2}B}=\frac{sin(A-B)}{sin(A+B)}[fromEq.(i\left)\right]$

$\Rightarrow \frac{si{n}^{2}A-si{n}^{2}B}{si{n}^{2}A+si{n}^{2}B}=\frac{sin(A-B)}{sin(A+B)}$

$\Rightarrow \frac{sin(A+B)sin(A-B)}{si{n}^{2}A+si{n}^{2}B}=\frac{sin(A-B)}{sin(A+B)}$

$\Rightarrow \frac{sin(A+B)sin(A-B)}{si{n}^{2}A+si{n}^{2}B}-\frac{sin(A-B)}{sin(A+B)}=0$

$\Rightarrow sin(A-B)[\frac{sin(A-B)}{si{n}^{2}A+si{n}^{2}B}-\frac{1}{sin(A+B)}]=0$

$\Rightarrow Eithersin(A-B)=0$

OR

$si{n}^2(A+B)=si{n}^{2}A+si{n}^{2}B$

$\Rightarrow A-B=0orsi{n}^{2}C=si{n}^{2}A+si{n}^{2}B$

$[\because sin(A+B)=sinC]$

$\Rightarrow A=Bor\frac{c^2}{k^2}=\frac{a^2}{k^2}+\frac{b^2}{k^2}[fromEq,(i\left)\right]$

$\Rightarrow A=Bor{a}^2+b^2=c^2$

Hence,it is either triangle is isosceles or right-angled.