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.

COMMENT.- First note that if $C=90^{\circ}$ then you have $\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\circ}$ you can consider that $\sin C=\sin (180^{\circ}-(A+B))=\sin(A+B)$ and $\cos C= \cos (180^{\circ}-(A+B))=-\cos(A+B)$ so you get
$$\sin^2 A+\sin^2 B+\sin^2 (A+B)=2(\cos^2A+\cos ^2 B+\cos^2(A+B))$$
You have to know simple formulas of trigonometry to finish.
EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\sin^2 A+\sin^2 B+\sin^2 (A+B) = 2\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\dfrac{\pi}{2}$ (the "if" is obvious and we need the "only if"). So put $A + B = \dfrac{\pi}{2}\pm h$ with $h\ne 0$ and look at the functions
$$\sin^2(x)+\sin^2(\dfrac{\pi}{2}\pm h-x)+\sin^2(\dfrac{\pi}{2}\pm h),\quad0\lt x\lt\dfrac{\pi}{2}$$ which becomes for $h$ positive and $h$ negative respectively $$f_1(x)=\sin^2(x)+\cos^2(x-h)+\cos^2(h)\qquad (1)\\f_2(x)=\sin^2(x)+\cos^2(x+h)+\cos^2(h)\qquad (2)$$ Now you can prove that $$\begin{equation}f_1(x)\begin{cases}\lt2\text{ when } 0\lt x\lt h\\=2 \text { when }x=h \space\text {discarded because } B\ne\dfrac{\pi}{2} \\\gt 2\text { when }h\lt x\lt\dfrac{\pi}{2}\\ \end{cases}\end{equation}$$
and that $f_2(x)$ is   smaller than $2$ on $0\lt x\lt \dfrac{\pi}{2}$. 
This requires elementary calculus that I guess you know how to handle.
