Solving problem with some basic trigonometry formula and factor formula

 ABC is a triangle such that $ \displaystyle {{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2$ ,
show that $ \displaystyle \Delta \ ABC$ is a right triangle .

$ \displaystyle {{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\ \ ......(1)$

$ \displaystyle {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1\ ......(2)$

$ \displaystyle (2)-(1)\Rightarrow $

$ \displaystyle {{\cos }^{2}}A-{{\sin }^{2}}A+{{\cos }^{2}}B-{{\sin }^{2}}B+{{\cos }^{2}}C-{{\sin }^{2}}C=-1$

$ \displaystyle \cos 2A+\cos 2B+\cos 2C=-1$

$ \displaystyle 2\cos (A+B)\cos (A-B)+2{{\cos }^{2}}C-1=-1$

$ \displaystyle 2\cos ({{180}^{\circ }}-C)\cos (A-B)+2{{\cos }^{2}}C=0$

$ \displaystyle -2\cos C\cos (A-B)+2{{\cos }^{2}}C=0$

$ \displaystyle 2\cos C\left[ {\cos C-\cos (A-B)} \right]=0$

$ \displaystyle -4\cos C\sin \frac{{C+A-B}}{2}\sin \frac{{C+B-A}}{2}=0$

$ \displaystyle \cos C\sin \frac{{{{{180}}^{\circ }}-2B}}{2}\sin \frac{{{{{180}}^{\circ }}-2A}}{2}=0$

$ \displaystyle \cos C\sin ({{90}^{\circ }}-B)\sin ({{90}^{\circ }}-A)=0$

$ \displaystyle \cos C\cos B\cos A=0$

$ \displaystyle \cos A=0\ \ \ (\text{or})\ \ \ \cos B=0\ \ \ (\text{or})\ \ \ \cos C=0$

$ \displaystyle \ \ \ \ \ A={{90}^{\circ }}\ (\text{or})\ \ \ \ \ B={{90}^{\circ }}\ \ (\text{or})\ \ \ \ \ \ \ \ C={{90}^{\circ }}$

$ \displaystyle \therefore \Delta \ ABC\ $ is a right triangle .