$\sin \left({\tan }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{9}-{\tan }^{-1}\frac{1}{7}\right)=$

$\sin \left({\tan }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{9}-{\tan }^{-1}\frac{1}{7}\right)$

$=\sin \left({\tan }^{-1}\frac{4}{5}-{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{9}\right)$

Now, ${\tan }^{-1}\frac{4}{5}-{\tan }^{-1}\frac{1}{7}=p\left(\mathrm{say}\right)$

$\Rightarrow p={\tan }^{-1}\left(\frac{{\displaystyle \frac{4}{5}}-{\displaystyle \frac{1}{7}}}{1+{\displaystyle \frac{4}{5}}·{\displaystyle \frac{1}{7}}}\right)={\tan }^{-1}\left(\frac{23}{39}\right)$

Also, ${\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{9}=q\left(\mathrm{say}\right)$

$\Rightarrow q={\tan }^{-1}\left(\frac{{\displaystyle \frac{4}{3}}+{\displaystyle \frac{1}{9}}}{1-{\displaystyle \frac{4}{3}}·{\displaystyle \frac{1}{9}}}\right)={\tan }^{-1}\left(\frac{39}{23}\right)$

$\Rightarrow q={\cot }^{-1}\left(\frac{23}{39}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\frac{23}{39}\right)$

So, $\sin \left(p+q\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right)=1$