If p = tan 20^{circ},then the value of frac{t | vedclass

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(C) Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have: $\frac{	an 160^{\circ} - 	an 110^{\circ}}{1 + 	an 160^{\ci

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$\frac{1-p^2}{2p}$

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(C) Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have: $\frac{	an 160^{\circ} - 	an 110^{\circ}}{1 + 	an 160^{\circ} 	an 110^{\circ}} = 	an(160^{\circ} - 110^{\circ}) = 	an 50^{\circ}$ Since $	an 50^{\circ} = 	an(90^{\circ} - 40^{\circ}) = \cot 40^{\circ} = \frac{1}{	an 40^{\circ}}$ Using the double angle formula $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta}$,we substitute $	heta = 20^{\circ}$: $	an 40^{\circ} = \frac{2	an 20^{\circ}}{1 - 	an^2 20^{\circ}} = \frac{2p}{1 - p^2}$ Therefore,$\frac{1}{	an 40^{\circ}} = \frac{1 - p^2}{2p}$.

If $p = \tan 20^{\circ}$,then the value of $\frac{\tan 160^{\circ} - \tan 110^{\circ}}{1 + \tan 160^{\circ} \tan 110^{\circ}}$ in terms of $p$ is

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