Given that $\alpha \& \beta$ root of

$3x^2-16x+5=0$

We know that

$\alpha +\beta =-\frac{b}{a}$

$\alpha \beta =\frac{c}{a}$, where $a=3, b=-16 \& c=5$

Now $\alpha +\beta =-\left(\frac{-16}{3}\right)=\frac{16}{3}$

$\alpha ·\beta =\frac{5}{3}$

if $xy>1$, then ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)+\mathrm{\pi }$

Where $\alpha ·\beta =\frac{5}{3}>1$

So,

${\tan }^{-1}\alpha +{\tan }^{-1}\beta -{\tan }^{-1}\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)=\mathrm{\pi }+{\tan }^{-1}\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)-{\tan }^{-1}\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)$

$\Rightarrow \mathrm{\pi }$