4.3. If \( 4 \tan \theta+3=0 \) and \( \theta \in\left[90^{\circ} ; 270^{\circ}\right] \), determine with the aid of a diagram, and without the use of a calculator, the value of : 4.3.1. \( \cos \theta \) 4.3.2. \( 1-\sin ^{2} \theta \)

To solve the equation \( 4 \tan \theta + 3 = 0 \), we can start by isolating the tangent: \[ \tan \theta = -\frac{3}{4} \] Since \( \theta \) is in the interval \( [90^{\circ}, 270^{\circ}] \), this places \( \theta \) in either the second quadrant or the third quadrant, where tangent is negative. To visualize this, draw the unit circle and notice how the tangent function relates to the sine and cosine ratios. 4.3.1. In the second quadrant, we can express \( \tan \theta \) as \( \frac{\sin \theta}{\cos \theta} = -\frac{3}{4} \). Let's denote \( \sin \theta = -3k \) and \( \cos \theta = 4k \). The identity \( \sin^2 \theta + \cos^2 \theta = 1 \) allows us to find \( k \): \[ (-3k)^2 + (4k)^2 = 1 \Rightarrow 9k^2 + 16k^2 = 1 \Rightarrow 25k^2 = 1 \Rightarrow k^2 = \frac{1}{25} \Rightarrow k = \frac{1}{5} \] Thus \( \sin \theta = -\frac{3}{5} \) and \( \cos \theta = 4k = \frac{4}{5} \) will give us our first answer. Since we need \( \cos \theta \): **Answer:** \( \cos \theta = \frac{4}{5} \) 4.3.2. Now, to find \( 1 - \sin^2 \theta \), we use the sine value we found earlier. We know: \[ \sin^2 \theta = \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \] This gives us: \[ 1 - \sin^2 \theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \] So: **Answer:** \( 1 - \sin^2 \theta = \frac{16}{25} \)