To solve the problem, we start with the given information: \(\sin x = -\frac{3}{5}\), where \(\pi < x < \frac{3\pi}{2}\). This range indicates that \(x\) is in the third quadrant, where sine is negative, and both cosine and tangent are also negative.
\(\sin^2 x + \cos^2 x = 1\)
\(\left(-\frac{3}{5}\right)^2 + \cos^2 x = 1\)
\(\frac{9}{25} + \cos^2 x = 1\)
\(\cos^2 x = 1 - \frac{9}{25}\)
\(\cos^2 x = \frac{16}{25}\)
\(\cos x = -\frac{4}{5}\) (since cosine is negative in the third quadrant)
\(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}}\)
\(\tan x = \frac{3}{4}\)
\(\tan^2 x = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
\(\tan^2 x - \cos x = \frac{9}{16} - \left(-\frac{4}{5}\right)\)
\(\frac{9}{16} = \frac{45}{80}\) and \(\left(-\frac{4}{5}\right) = -\frac{64}{80}\)
\(\tan^2 x - \cos x = \frac{45}{80} + \frac{64}{80} = \frac{109}{80}\)
\(80 \times \frac{109}{80} = 109\)
This is the correct option among the given choices, justifying that \(80(\tan^2 x - \cos x)\) evaluates to 109 in the specified conditions.
Given:
\[ \sin x = -\frac{3}{5}, \quad \pi < x < \frac{3\pi}{2}. \]
Step 1: Use the Pythagorean identity:
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}. \]
Step 2: Determine \( \cos x \):
Since \( \cos x < 0 \) in the third quadrant:
\[ \cos x = -\frac{4}{5}. \]
Step 3: Calculate \( \tan x \):
\[ \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}. \]
Step 4: Compute \( 80(\tan^2 x - \cos x) \):
\[ \tan^2 x = \left(\frac{3}{4}\right)^2 = \frac{9}{16}, \quad 80\left(\tan^2 x - \cos x\right) = 80\left(\frac{9}{16} - \left(-\frac{4}{5}\right)\right). \]
Step 5: Simplify:
\[ 80\left(\frac{9}{16} + \frac{4}{5}\right) = 80\left(\frac{45}{80} + \frac{64}{80}\right) = 80 \cdot \frac{109}{80} = 109. \]
Final Answer:
\[ \boxed{109.} \]
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to