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

Ask by Ingram Bernard. in South Africa

Feb 21,2025

To solve the problem, we start with the given condition: \[ \sin \theta = \frac{-6k}{k^2 + 9} \] We know that \(\theta\) is in the interval \((90^\circ, 270^\circ)\), which means \(\sin \theta\) is negative and \(\cos \theta\) will also be negative in this range. ### Step 1: Determine \(\cos \theta\) Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] We can express \(\cos^2 \theta\) in terms of \(\sin \theta\): \[ \cos^2 \theta = 1 - \sin^2 \theta \] Substituting \(\sin \theta\): \[ \cos^2 \theta = 1 - \left(\frac{-6k}{k^2 + 9}\right)^2 \] Calculating \(\sin^2 \theta\): \[ \sin^2 \theta = \frac{36k^2}{(k^2 + 9)^2} \] Now substituting this into the equation for \(\cos^2 \theta\): \[ \cos^2 \theta = 1 - \frac{36k^2}{(k^2 + 9)^2} \] To combine these terms, we need a common denominator: \[ \cos^2 \theta = \frac{(k^2 + 9)^2 - 36k^2}{(k^2 + 9)^2} \] Now we simplify the numerator: \[ (k^2 + 9)^2 - 36k^2 = k^4 + 18k^2 + 81 - 36k^2 = k^4 - 18k^2 + 81 \] Thus, we have: \[ \cos^2 \theta = \frac{k^4 - 18k^2 + 81}{(k^2 + 9)^2} \] Since \(\theta\) is in the second quadrant, \(\cos \theta\) is negative: \[ \cos \theta = -\sqrt{\frac{k^4 - 18k^2 + 81}{(k^2 + 9)^2}} = -\frac{\sqrt{k^4 - 18k^2 + 81}}{k^2 + 9} \] ### Step 2: Determine \(\tan \theta + \frac{1}{\cos \theta}\) We know that: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] Substituting the expressions we found: \[ \tan \theta = \frac{\frac{-6k}{k^2 + 9}}{-\frac{\sqrt{k^4 - 18k^2 + 81}}{k^2 + 9}} = \frac{6k}{\sqrt{k^4 - 18k^2 + 81}} \] Now we can find \(\tan \theta + \frac{1}{\cos \theta}\): \[ \frac{1}{\cos \theta} = -\frac{k^2 + 9}{\sqrt{k^4 - 18k^2 + 81}} \] Thus: \[ \tan \theta + \frac{1}{\cos \theta} = \frac{6k}{\sqrt{k^4 - 18k^2 + 81}} - \frac{k^2 + 9}{\sqrt{k^4 - 18k^2 + 81}} \] Combining these fractions: \[ \tan \theta + \frac{1}{\cos \theta} = \frac{6k - (k^2 + 9)}{\sqrt{k^4 - 18k^2 + 81}} = \frac{6k - k^2 - 9}{\sqrt{k^4 - 18k^2 + 81}} \] ### Final Results 1. \(\cos \theta = -\frac{\sqrt{k^4 - 18k^2 + 81}}{k^2 + 9}\) 2. \(\tan \theta + \frac{1}{\cos \theta} = \frac{6k - k^2 - 9}{\sqrt{k^4 - 18k^2 + 81}}\)