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If $p=\tan 20^{\circ}$, then 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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$\frac{1-p^2}{2 p}$
$\begin{aligned} & \frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \cdot \tan 110^{\circ}}=\tan \left(160^{\circ}-110^{\circ}\right)=\tan 50^{\circ} \\ & =\tan \left(90^{\circ}-40^{\circ}\right)=\cot 40^{\circ}=\frac{1}{\tan 40^{\circ}} \\ & =\frac{1}{\frac{2 \tan _2 0^{\circ}}{1-\tan ^2 20^{\circ}}}=\frac{1-\tan ^2 20^{\circ}}{2 \tan 20^{\circ}}=\frac{1-p^2}{2 p}\end{aligned}$
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