For the following two (02) items: Let p = cosec\, 20° and q =...

Given p = \text{cosec}\, 20^{}^{\circ} = \dfrac{1}{\sin 20^{}^{\circ}} and q = \text{cosec}\, 70^{}^{\circ} = \dfrac{1}{\sin 70^{}^{\circ}} = \dfrac{1}{\cos 20^{}^{\circ}}

Substituting $p$ and $q$ into the given expression:

\dfrac{\sqrt{3}\,p}{4} - \dfrac{q}{4} = \dfrac{\sqrt{3}}{4 \sin 20^{}^{\circ}} - \dfrac{1}{4 \cos 20^{}^{\circ}}

Taking the common denominator:

= \dfrac{\sqrt{3} \cos 20^{}^{\circ} - \sin 20^{}^{\circ}}{4 \sin 20^{}^{\circ} \cos 20^{}^{\circ}}

Multiplying and dividing the numerator by $2$, and rewriting the denominator:

= \dfrac{2 \left( \dfrac{\sqrt{3}}{2} \cos 20^{}^{\circ} - \dfrac{1}{2} \sin 20^{}^{\circ} \right)}{2 (2 \sin 20^{}^{\circ} \cos 20^{}^{\circ})}

Using the identities \sin 60^{}^{\circ} = \dfrac{\sqrt{3}}{2}, \cos 60^{}^{\circ} = \dfrac{1}{2}, and $\sin 2\theta = 2 \sin \theta \cos \theta$:

= \dfrac{2 (\sin 60^{}^{\circ} \cos 20^{}^{\circ} - \cos 60^{}^{\circ} \sin 20^{}^{\circ})}{2 \sin 40^{}^{\circ}}

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

= \dfrac{2 \sin(60^{}^{\circ} - 20^{}^{\circ})}{2 \sin 40^{}^{\circ}}

= \dfrac{2 \sin 40^{}^{\circ}}{2 \sin 40^{}^{\circ}}

$= 1$

Answer: $1$