1. Given that $PQ$ is a diameter of the circle, $PR$ and $QS$ are chords intersecting at $T$, and $\angle PTQ = \theta$.
2. We need to determine the ratio of the area of $\triangle QTP$ to the area of $\triangle SRT$ in terms of trigonometric functions of $\theta$.

First, observe that $\triangle PQT \sim \triangle SRT$ because:
- $\angle TPQ = \angle RPQ = \angle RSQ = \angle RST$ (angles subtended by the same arc are equal).
- $\angle TQP = \angle SQP = \angle SRP = \angle SRT$ (angles subtended by the same arc are equal).

Since $\triangle PQT \sim \triangle SRT$, the ratio of their areas is the square of the ratio of their corresponding sides. Let $\frac{TP}{TS} = r$. Then, the desired ratio of the areas is $r^2$.

Next, observe that $\angle PST = \angle PSQ = 90^\circ$ because $PQ$ is a diameter (by the Inscribed Angle Theorem, an angle subtended by a diameter is a right angle).

Now, consider the triangle $\triangle PST$:
- $\angle STP = 180^\circ - \theta$ (since $\angle PTQ = \theta$ and $\angle PTQ$ and $\angle STP$ are supplementary).

Using the cosine rule in $\triangle PST$:
$$\frac{TP}{TS} = \frac{1}{\cos(\angle STP)} = \frac{1}{\cos(180^\circ - \theta)}$$

Since $\cos(180^\circ - \theta) = -\cos(\theta)$, we have:
$$\frac{TP}{TS} = \frac{1}{-\cos(\theta)} = -\frac{1}{\cos(\theta)}$$

However, since $\theta > 90^\circ$, $\cos(\theta)$ is negative, making $-\frac{1}{\cos(\theta)}$ positive. Therefore:
$$\frac{TP}{TS} = \frac{1}{|\cos(\theta)|}$$

The desired ratio of the areas is:
$$\left( \frac{TP}{TS} \right)^2 = \left( \frac{1}{|\cos(\theta)|} \right)^2 = \frac{1}{\cos^2(\theta)}$$

The final answer is $\boxed{\frac{1}{\cos^2(\theta)}}$