If ${\tan }^2\frac{\pi -A}{4}+{\tan }^2\frac{\pi -B}{4}+{\tan }^2\frac{\pi -C}{4}=1$ , then $\Delta ABC$ is

${\tan }^2\alpha +{\tan }^2\beta +{\tan }^2\gamma =1\to \left(1\right)$

$\alpha +\beta +\gamma =\frac{\pi }{2}+\frac{\pi }{4}-\frac{1}{4}\left(A+B+C\right)$

$\theta =\frac{\pi }{2}+\frac{\pi }{4}-\frac{1}{4}\left(\pi \right)$

$=\frac{\pi }{2}+\frac{\pi }{4}-\frac{\pi }{4}$

$=\frac{\pi }{2}$

$\Rightarrow \alpha +\beta =\frac{\pi }{2}-\gamma$

$\Rightarrow \tan \left(\alpha +\beta \right)=\tan \left(\frac{\pi }{2}-\gamma \right)=\cot \gamma =\frac{1}{\tan \gamma }$

$\Rightarrow \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1}{\tan \gamma }$

$\Rightarrow \tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =1\to \left(2\right)$

${\tan }^2\alpha {\tan }^2\beta +{\tan }^2\gamma =1=\tan \alpha \tan \beta +\tan \beta \tan \gamma +\tan \gamma \tan \alpha =1$

$\iff \tan \alpha =\tan \beta =\tan \gamma$

$\iff \alpha =\beta =\gamma$

$\iff \frac{\pi -A}{4}=\frac{\pi -B}{4}=\frac{\pi -C}{4}$

$\Rightarrow A=B=C=60°$