P(-1; 2), R(3; 0) and Q(-3; a) are the vertices of a triangle in the Cartesian plane and PR = 1 Determine the coordinates of E, the midpoint of PR 2 Show that a = -2 if the midpoint of QR is (0; -1) 3 Determine the gradient of EQ

In the right-angled triangle EFG, with $\angle E = 90^\circ$ and $\angle F = \theta$:
The side opposite to $\theta$ is EG = 2 units.
The side adjacent to $\theta$ is EF = 3 units.

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$.

The value of $\tan \theta$ is $\boxed*{\frac{2}{3}*}$.

2.1.2 Write down the value of $\csc \theta$.

The side opposite to $\theta$ is EG = 2 units.
The hypotenuse is FG = $\sqrt{13}$ units.

$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$.

$\csc \theta = \frac{1}{\sin \theta}$.

The value of $\csc \theta$ is $\boxed*{\frac{\sqrt{13}}{2}*}$.

2.1.3 Write down the value of $\cos(90^\circ - \theta)$.

Since $\angle E = 90^\circ$ and $\angle F = \theta$, the third angle $\angle G = 180^\circ - 90^\circ - \theta = 90^\circ - \theta$.

Therefore, $\cos(90^\circ - \theta) = \frac{2}{\sqrt{13}}$.
The value of $\cos(90^\circ - \theta)$ is $\boxed*{\frac{2}{\sqrt{13}}*}$.