<s>[INST] If $ \tan(20^{\circ}) = p $, find $ \frac{\tan(160^{\circ}) - \tan(110^{\circ})}{1 + \tan(160^{\circ}) \tan(110^{\circ})} $. 
I applied the $ \tan(A - B) $-formula to make it $\tan(50^{\circ}) $, then I split it to $ \tan(30^{\circ} + 20^{\circ}) $. My answer came out to be $ \dfrac{\sqrt[3]{p + 1}}{\sqrt[3]{- p}} $, but the actual answer is $ \dfrac{1 - p^{2}}{2 p} $.
 [/INST] $$\frac{\tan 160 - \tan 110}{1+\tan 160 \tan 110} = \tan(160 - 110) = \tan 50 = \cot 40 = \frac{1}{\tan 40} \\
\tan 40 = \frac{\tan 20 + \tan 20}{1-\tan^2 20} = \frac{2p}{1-p^2} \\
\frac{\tan 160 - \tan 110}{1+\tan 160 \tan 110} =  \frac{1-p^2}{2p}
$$ 
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