In the diagram, D, B and C are points in the same horizontal plane. AB is a vertical flagpole. A rope is tied from A, the top of the flagpole, to point C. BAC = BCD = θ, BDC = α and AB = 4 units.

Let H be the foot of the perpendicular from R to TP. So, RH is the altitude.
In right-angled $\triangle RHT$:

Given $RT = p$, so:

In right-angled $\triangle RHP$:

Given $RP = t$, so:

From Equation 1 and Equation 2:

Divide both sides by $pt$:

This proves the identity.

8.2.1 Show that the distance between B and C is $4 \tan \theta$.

Consider $\triangle ABC$. AB is a vertical flagpole and B, C are points in the horizontal plane, so $\angle ABC = 90^\circ$.
We are given $AB = 4$ units and $\text{B}\hat{\text{A}}\text{C} = \theta$.

In right-angled $\triangle ABC$:

Substitute the given values:

This shows that the distance between B and C is $4 \tan \theta$.

8.2.2 Hence, show that $\text{DC} = 4 \sin \theta \left(1 + \frac{\tan \theta}{\tan \alpha}\right)$.

We have $BC = 4 \tan \theta$ from 8.2.1.
In $\triangle BCD$, we are given $\text{B}\hat{\text{C}}\text{D} = \theta$ and $\text{B}\hat{\text{D}}\text{C} = \alpha$.
The sum of angles in a triangle is $180^\circ$, so $\text{D}\hat{\text{B}}\text{C} = 180^\circ - (\theta + \alpha)$.
Using the Sine Rule:

Since $\sin(180^\circ - X) = \sin X$:

Substitute $BC = 4 \tan \theta$:

Recall $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \alpha = \frac{1}{\tan \alpha}$.

Factor out $\cos \theta$ from the bracket:

This shows the required expression for DC.