If \( \tan \theta=\frac{12}{5} \) and \( 90^{\circ}

Ask by Daniel Owen. in South Africa

Jan 21,2025

Given that \( \tan \theta = \frac{12}{5} \), we can find the corresponding values of sine and cosine. Using the definition of tangent: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{12}{5} \] Let's set \( \sin \theta = 12k \) and \( \cos \theta = 5k \) for some positive value \( k \). Then, using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ (12k)^2 + (5k)^2 = 1 \] Calculating the squares: \[ 144k^2 + 25k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13} \] Now, substituting \( k \) back to find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = 12k = \frac{12}{13}, \quad \cos \theta = 5k = \frac{5}{13} \] Considering the range \( 90^\circ < \theta < 270^\circ \), which means \( \theta \) is in the second or third quadrant, we note that cosine is negative in both quadrants, while sine is positive in the second quadrant and negative in the third quadrant. Therefore: - In this case, for \( \theta \) in the third quadrant, \( \sin \theta = -\frac{12}{13} \) and \( \cos \theta = -\frac{5}{13} \). Now, we need to compute: \[ \frac{10}{\cos \theta} - \frac{5}{\cos \theta} \] Factoring out \( \frac{1}{\cos \theta} \): \[ \frac{10 - 5}{\cos \theta} = \frac{5}{\cos \theta} \] Substituting \( \cos \theta = -\frac{5}{13} \): \[ \frac{5}{\cos \theta} = \frac{5}{-\frac{5}{13}} = 5 \cdot -\frac{13}{5} = -13 \] Thus, the value of \( \frac{10}{\cos \theta} - \frac{5}{\cos \theta} \) is \[ \boxed{-13} \]