Question

In $\Delta ABC,$ if ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C,$ then show that the triangle is a right angled triangle.

Answer 1

${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$

${\sin }^{2}A={\sin }^{2}C-{\sin }^{2}B$

${\sin }^{2}A=\sin (C-B)\sin (C+B)$

as $\sin A=\sin (\pi -(B+C))=\sin (B+C)$

${\sin }^2(B+C)=\sin (C-B)\sin (C+B)$

$\sin (C+B)=\sin (C-B)$

Thus $C+B=\pi -(C-B);C+B\ne C-B$ because then $B$ will become $0$

$C=\frac{\pi }{2}$

Therefore triangle is right angled

$\sin (\alpha -\beta )(\sin \alpha +\beta )$

$(\sin \alpha \cos \beta {)}^2$ -$(\sin \beta -\cos \alpha {)}^2$

${\sin }^2\alpha -{\sin }^2\beta$