Question

If $\sin x = -\frac{3}{5}$, where $\pi<x<\frac{3\pi}{2}$, then $80(\tan^2 x - \cos x)$ is equal to:

• A. 109
• B. 108
• C. 18
• D. 19

Answer 1

Answer: (A)

109

Answer 2

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:

$\boxed{109.}$