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:
\[ \boxed{109.} \]
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
