Question

In $△ABC$, if $\angle C=\frac{\pi }{2}$, then prove that $\sin (A-B)=\frac{a^2-b^2}{a^2+b^2}$.

Answer 1

$A+B+C=\pi$

$\Rightarrow A+B=\frac{\pi }{2}\therefore B=\frac{\pi }{2}-A\sin (A-B)=\sin (A-\frac{\pi }{2}+A)=\sin (2A-\frac{\pi }{2})=-\sin (\frac{\pi }{2}-2A)=-\cos \left(2A\right)=-(2{cos}^{2}A-1)=1-2{cos}^{2}A$

$\cos A=\frac{b}{c}\therefore 1-2{\cos }^{2}A=1-2\left(\frac{b^2}{c^2}\right)=\frac{c^2-2b^2}{c^2}$

By Pythagoras theorem,

$c^2=a^2+b^2⟶1\Rightarrow \sin (A-B)=1-2{\cos }^{2}A=\frac{c^2-2b^2}{c^2}$
Using $1$

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