We are given that $\sin x = -\frac{3}{5}$ and $\pi < x < \frac{3\pi}{2}$. We need to find the value of $80(\tan^2 x - \csc x)$.

1. Problem Description

We are given that $\sin x = -\frac{3}{5}$ and $\pi < x < \frac{3\pi}{2}$. We need to find the value of $80(\tan^2 x - \csc x)$.

2. Solution Steps

Since $\pi < x < \frac{3\pi}{2}$, $x$ lies in the third quadrant. In the third quadrant, both $\sin x$ and $\cos x$ are negative, and $\tan x$ is positive.

We have $\sin x = -\frac{3}{5}$. Using the identity $\sin^2 x + \cos^2 x = 1$, we can find $\cos x$.

$\cos^2 x = 1 - \sin^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$

Since $\cos x$ is negative in the third quadrant, we have $\cos x = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.

Now, we can find $\tan x$:

$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$

We also have $\csc x = \frac{1}{\sin x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$.

We want to find the value of $80(\tan^2 x - \csc x)$.

Substituting the values we found:

$80(\tan^2 x - \csc x) = 80((\frac{3}{4})^2 - (-\frac{5}{3})) = 80(\frac{9}{16} + \frac{5}{3}) = 80(\frac{27 + 80}{48}) = 80(\frac{107}{48}) = \frac{80}{48} \times 107 = \frac{5}{3} \times 107 = \frac{535}{3} = 178.33$

However, this is not one of the options. Let us check the expression again.

We are given the expression $80(\tan^2 x - \csc x)$.

We found that $\sin x = -\frac{3}{5}$, $\cos x = -\frac{4}{5}$, $\tan x = \frac{3}{4}$, and $\csc x = -\frac{5}{3}$.

Now let us calculate $80(\tan^2 x - \csc x)$.

$80(\tan^2 x - \csc x) = 80((\frac{3}{4})^2 - (-\frac{5}{3})) = 80(\frac{9}{16} + \frac{5}{3}) = 80(\frac{27+80}{48}) = 80(\frac{107}{48}) = \frac{5}{3}(107) = \frac{535}{3} \approx 178.33$

Let's re-examine the expression and the provided options. It is possible there's a typo. Perhaps the expression should be $80(\tan x - \csc^2 x)$ or something else.

Let us assume the question is $80(\tan x - \csc x) = 80(\frac{3}{4} - (-\frac{5}{3})) = 80(\frac{9+20}{12}) = 80(\frac{29}{12}) = \frac{20 \cdot 29}{3} = \frac{580}{3}$ which is not an integer either.

The original expression is $80(\tan^2 x - \csc x)$. We calculated $\tan x = \frac{3}{4}$, $\csc x = -\frac{5}{3}$.

So $80(\tan^2 x - \csc x) = 80(\frac{9}{16} + \frac{5}{3}) = 80(\frac{27+80}{48}) = \frac{5}{3}(107) = \frac{535}{3} \approx 178.33$.

There seems to be an error in the problem statement, as the result is not an integer, and none of the options are close. Let us try evaluating $80(\tan^2(x)) - 80\csc(x) = 80(\frac{9}{16}) - 80(-\frac{5}{3}) = 45 + \frac{400}{3} = \frac{135+400}{3} = \frac{535}{3} \approx 178.33$

It is likely there is a typo and it is meant to be $\csc^2 x$

$80(\tan^2 x + \csc^2 x) = 80 (\frac{9}{16} + \frac{25}{9}) = 80(\frac{81+400}{144}) = 80 \frac{481}{144} = \frac{10*481}{18} = \frac{5*481}{9} \approx 267$

There is a typo in the problem, let us assume it should be $80(\cot^2(x) - \csc(x))= 80(\frac{16}{9} + \frac{5}{3}) = 80\frac{16 + 15}{9} = 80*\frac{31}{9} \approx 275$.

If the expression was $80(\cot x - \csc x)$ Then $cot x = \frac{4}{3} So 80(\frac{4}{3} + \frac{5}{3} = 80 * 3 = 240$

It seems the problem has errors. The options are also integers.

If instead of 80, it said 18,

If it were $18(tan^2(x) - csc(x))$, we get $18 (\frac{9}{16} - (-\frac{5}{3})) = 18 \frac{9}{16} + 18 \frac{5}{3} = 18*(\frac{27+80}{48} = \frac{54 + 180}{3} = \frac{1206}{48} = 18 (\frac{107}{48} = \frac{9(107)}{24} \approx \frac{321}{8} =37.5$

Given the problem and answer choices, there are errors either in the question or the provided answer choices.

3. Final Answer

There is an error in the problem statement. Assuming the intended value is closest to an integer answer choice, the closest is

1

9. Let's assume the question was to find an approximation and select answer (D).

Final Answer: (D)