If $2\tan^2A\tan^2B\tan^2C + \tan^2A\tan^2B + \tan^2B\tan^2C + \tan^2C\tan^2A = 1$, then find the value of $\sin^2A + \sin^2B + \sin^2C$.
My attempt
1). I tried to multiply both sides by $\cos^2A\cos^2B\cos^2C$ in given. Then took common but after that, I didn't get what to do next
2). I tried to relate the given with $(ab+bc+ca)^2,$ but it also didn't work... I could only think of these two ways but nothing worked for me...
$$\sum_{cyc}\sin^2\alpha=\sum_{cyc}\frac{\tan^2\alpha}{1+\tan^2\alpha}=\frac{\sum\limits_{cyc}\tan^2\alpha(1+\tan^2\beta)(1+\tan^2\gamma)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=$$ $$=\frac{\sum\limits_{cyc}(\tan^2\alpha\tan^2\beta\tan^2\gamma+2\tan^2\alpha\tan^2\beta+\tan^2\alpha)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=$$ $$=\frac{\tan^2\alpha\tan^2\beta\tan^2\gamma+1+\sum\limits_{cyc}(\tan^2\alpha\tan^2\beta+\tan^2\alpha)}{\prod\limits_{cyc}(1+\tan^2\alpha)}=1.$$