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 one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: