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

Question

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

Aicharya · Accepted Answer

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

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

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