If $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$ , then $\triangle ABC$ is a right triangle 
Show that the triangle whose angles satisfy the equality
  $$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$
  is right-angled.

I've tried many times, but was unsuccessful.

\begin{align}
\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}
&=2
. 
\end{align}  
There are known identities 
for any $\triangle ABC$
with sides $a,b,c$, angles $A,B,C$
radius of inscribed circle $r$,
radius of circumscribed circle $R$
and semiperimeter $\rho=\tfrac12(a+b+c)$:
\begin{align}
a^2+b^2+c^2
&=
2\rho^2-8r\,R-2r^2
,\\
\cos A \cos B \cos C
&=\frac{\rho^2-(r+2R)^2}{4R^2}
.
\end{align} 
\begin{align}
{\sin^2 A + \sin^2 B + \sin^2 C}
&=2(\cos^2 A + \cos^2 B + \cos^2 C)
,\\
{\sin^2 A + \sin^2 B + \sin^2 C}
&=2(1-\sin^2 A + 1-\sin^2 B + 1-\sin^2 C)
,\\
3(\sin^2 A + \sin^2 B + \sin^2 C)
&=6
,\\
\sin^2 A + \sin^2 B + \sin^2 C
&=2
,\\
4R^2\sin^2 A + 4R^2\sin^2 B + 4R^2\sin^2 C
&=2\cdot4R^2
,\\
a^2+b^2+c^2&=8R^2
,\\
2\rho^2-8r\,R-2r^2&=8R^2
,\\
\rho^2-(r+2R)^2&=0
,\\
\frac{\rho^2-(r+2R)^2}{4R^2}
=0
.
\end{align}  
And since
\begin{align} 
\cos A \cos B \cos C
&=\frac{\rho^2-(r+2R)^2}{4R^2}
,
\end{align}
one of 
$\cos A,\cos B,\cos C$
must be 0,
hence, one of the angles must be $\tfrac\pi2$.
