If $\tan 20°=P$ then $\frac{\tan 160°-\tan 110°}{1+\tan 160°\tan 110°}=$

$=\frac{-\tan 20°+\cot 20°}{1+\tan 20°.\cot 20°}$

$=\frac{-p+\frac{1}{p}}{1+1}=\frac{-p^2+1}{2P}=\frac{1-p^2}{2p}$

$\Rightarrow$ $\boxed{\frac{\tan 160-\tan 110}{1+\tan 160.\tan 110}=\frac{-p^2+1}{2P}}$