Do not use a calculator in this question. Given that $\tan\theta=\frac{3}{5}$ and that $\theta$ is non-acute, find the exact value of

1. Find the value of $\cot\theta$

$\cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

2. Find the value of $\sec^2\theta$ using the Pythagorean identity

$\sec^2\theta = 1 + \tan^2\theta = 1 + \left(\frac{3}{5}\right)^2 = 1 + \frac{9}{25} = \frac{34}{25}$

3. Evaluate the given expression

$\cot\theta + \sec^2\theta = \frac{5}{3} + \frac{34}{25} = \frac{125}{75} + \frac{102}{75} = \frac{227}{75}$

1. Determine the quadrant in which $\theta$ lies

Since $\tan\theta = \frac{3}{5} > 0$ and $\theta$ is non-acute, $\theta$ must lie in the 3rd quadrant.

2. Calculate the hypotenuse of the reference triangle

Let the opposite side be $3$ and the adjacent side be $5$.
Hypotenuse $= \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$

3. Evaluate $\sin\theta$ with the correct sign

In the 3rd quadrant, sine is negative.
$\sin\theta = -\frac{\text{opposite}}{\text{hypotenuse}} = -\frac{3}{\sqrt{34}}$

1. Determine the quadrant of $\theta$ and find $\cos\theta$

Since $\tan\theta = \frac{3}{5} > 0$ and $\theta$ is non-acute, $\theta$ must lie in the third quadrant.
The hypotenuse is $\sqrt{3^2 + 5^2} = \sqrt{34}$.
In the third quadrant, cosine is negative, so $\cos\theta = -\frac{5}{\sqrt{34}}$.

2. Evaluate $\cos(180^\circ - \theta)$ using the appropriate identity

Using the identity $\cos(180^\circ - \theta) = -\cos\theta$:
$\cos(180^\circ - \theta) = -\left(-\frac{5}{\sqrt{34}}\right) = \frac{5}{\sqrt{34}}$

1. Simplify the given expression using negative angle and reciprocal identities

$\dfrac{\operatorname{cosec}(-\theta)}{\sec(\theta)} = \dfrac{-\operatorname{cosec}(\theta)}{\sec(\theta)}$

$= \dfrac{-\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}$

$= -\dfrac{\cos\theta}{\sin\theta}$

$= -\cot\theta$

2. Evaluate the simplified expression using the given value of $\tan\theta$

Since $\tan\theta = \frac{3}{5}$, we have $\cot\theta = \frac{1}{\tan\theta} = \frac{5}{3}$.

Therefore, $-\cot\theta = -\frac{5}{3}$.