If \( \sin \theta=\frac{-6 k}{k^{2}+9}, 0

Ask by Weaver Johnson. in South Africa

Mar 10,2025

Given that \( \sin \theta = \frac{-6 k}{k^{2} + 9} \) and \( \theta \in \left(90^{\circ}, 270^{\circ}\right) \), we note that in this interval, the sine function is negative, which matches our expression for \( \sin \theta \). ### Step 1: Determine \( \cos \theta \) We use the Pythagorean identity, which states \[ \sin^2 \theta + \cos^2 \theta = 1. \] Substituting in our expression for \( \sin \theta \): \[ \left( \frac{-6k}{k^{2} + 9} \right)^2 + \cos^2 \theta = 1. \] Calculating \( \left( \frac{-6k}{k^{2} + 9} \right)^2 \): \[ \frac{36k^2}{(k^2 + 9)^2}. \] Substituting this into the identity gives: \[ \frac{36k^2}{(k^2 + 9)^2} + \cos^2 \theta = 1 \] which leads to: \[ \cos^2 \theta = 1 - \frac{36k^2}{(k^2 + 9)^2}. \] Finding a common denominator: \[ \cos^2 \theta = \frac{(k^2 + 9)^2 - 36k^2}{(k^2 + 9)^2}. \] Now let's expand \( (k^2 + 9)^2 \): \[ (k^2 + 9)^2 = k^4 + 18k^2 + 81. \] Thus: \[ \cos^2 \theta = \frac{k^4 + 18k^2 + 81 - 36k^2}{(k^2 + 9)^2}, \] \[ \cos^2 \theta = \frac{k^4 - 18k^2 + 81}{(k^2 + 9)^2}. \] Recognizing the numerator as a perfect square: \[ k^4 - 18k^2 + 81 = (k^2 - 9)^2, \] we obtain: \[ \cos^2 \theta = \frac{(k^2 - 9)^2}{(k^2 + 9)^2}. \] Taking the square root, knowing \( \cos \theta < 0 \) in the specified quadrant: \[ \cos \theta = -\frac{k^2 - 9}{k^2 + 9}. \] ### Step 2: Determine \( \tan \theta + \frac{1}{\cos \theta} \) Using the definitions of tangent and previously calculated cosine, we find: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{-6k}{k^2 + 9}}{-\frac{k^2 - 9}{k^2 + 9}} = \frac{6k}{k^2 - 9}. \] Now, calculating \( \tan \theta + \frac{1}{\cos \theta} \): \[ \frac{1}{\cos \theta} = -\frac{k^2 + 9}{k^2 - 9}, \] therefore: \[ \tan \theta + \frac{1}{\cos \theta} = \frac{6k}{k^2 - 9} - \frac{k^2 + 9}{k^2 - 9}. \] Combining under a common denominator gives: \[ \tan \theta + \frac{1}{\cos \theta} = \frac{6k - (k^2 + 9)}{k^2 - 9} = \frac{6k - k^2 - 9}{k^2 - 9}. \] ### Final Answers Thus, the expressions in terms of \( k \) are \[ \cos \theta = -\frac{k^2 - 9}{k^2 + 9}, \quad \tan \theta + \frac{1}{\cos \theta} = \frac{6k - k^2 - 9}{k^2 - 9}. \]