Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.  
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 C)$
$3(\sin^2 A+\sin^2 B+\sin^2 C)=6$
$(\sin^2 A+\sin^2 B+\sin^2 C)=2$
What to do now? Please help!

From where you have left off,
$$\sin^2A+\sin^2B+\sin^2C=2\implies1-\cos^2A+1-(\cos^2B-\sin^2C)=2$$
$$\implies\cos^2A+\cos^2B-\sin^2C=0$$
Using $\displaystyle\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$
$$\implies\cos^2A+\cos^2B-\sin^2C=\cos^2A+\cos(B+C)\cos(B-C)$$
As $\displaystyle A+B+C=\pi, B+C=\pi-A,\cos(B+C)=\cos(\pi-A)=-\cos A,$
$$\implies\cos^2A+\cos(B+C)\cos(B-C)=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\left[\cos A -\cos(B-C)\right]$$
$$=\cos A\left[-\cos(B+C) -\cos(B-C)\right]$$
$$=\cos A\left[-2\cos B\cos C\right]$$
