To evaluate [tex]\(\tan \alpha + \sin \alpha\)[/tex] given that [tex]\(\operatorname{Car} x=\frac{3}{13}\)[/tex] and that [tex]\(270^{\circ} \leq \alpha \leq 360^{\circ}\)[/tex], follow these steps:
1. Understand the Problem and Given Information:
- We are given [tex]\(\operatorname{Car} x=\frac{3}{13}\)[/tex]. In trigonometry, "car" here appears to refer to [tex]\(\cos \alpha = \frac{3}{13}\)[/tex].
- [tex]\(\alpha\)[/tex] lies in the range between [tex]\(270^{\circ}\)[/tex] and [tex]\(360^{\circ}\)[/tex], placing [tex]\(\alpha\)[/tex] in the fourth quadrant.
2. Calculate [tex]\(\cos \alpha\)[/tex]:
[tex]\[\cos \alpha = \frac{3}{13}\][/tex]
3. Determine [tex]\(\sin \alpha\)[/tex]:
- We use the Pythagorean identity: [tex]\(\cos^2 \alpha + \sin^2 \alpha = 1\)[/tex].
- Substitute [tex]\(\cos \alpha\)[/tex]:
[tex]\[\cos^2 \alpha = \left(\frac{3}{13}\right)^2 = \frac{9}{169}\][/tex]
[tex]\[\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{9}{169} = \frac{160}{169}\][/tex]
- Since [tex]\(\alpha\)[/tex] is in the fourth quadrant, [tex]\(\sin \alpha\)[/tex] must be negative:
[tex]\[\sin \alpha = - \sqrt{\frac{160}{169}} = - \frac{\sqrt{160}}{13}\][/tex]
- Simplify [tex]\(\sin \alpha\)[/tex]:
[tex]\[\sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10}\][/tex]
[tex]\[\sin \alpha = - \frac{4\sqrt{10}}{13}\][/tex]
4. Calculate [tex]\(\tan \alpha\)[/tex]:
- Using the definition [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex]:
[tex]\[\tan \alpha = \frac{-\frac{4\sqrt{10}}{13}}{\frac{3}{13}} = -\frac{4\sqrt{10}}{3}\][/tex]
5. Evaluate [tex]\(\tan \alpha + \sin \alpha\)[/tex]:
- Substitute the values we found:
[tex]\[\tan \alpha = -\frac{4\sqrt{10}}{3}\][/tex]
[tex]\[\sin \alpha = -\frac{4\sqrt{10}}{13}\][/tex]
- Find the sum:
[tex]\[\tan \alpha + \sin \alpha = -\frac{4\sqrt{10}}{3} - \frac{4\sqrt{10}}{13}\][/tex]
6. Simplify the Expression:
- To combine these, find a common denominator:
[tex]\[\tan \alpha + \sin \alpha = -\frac{4\sqrt{10}}{3} - \frac{4\sqrt{10}}{13} = -\frac{4\sqrt{10} \cdot 13}{3 \cdot 13} - \frac{4\sqrt{10} \cdot 3}{13 \cdot 3} = -\frac{52\sqrt{10} + 12\sqrt{10}}{39}\][/tex]
[tex]\[\tan \alpha + \sin \alpha = -\frac{64\sqrt{10}}{39}\][/tex]
The evaluated values are:
[tex]\[
\cos \alpha = 0.23076923076923078, \quad \sin \alpha = -0.9730085108210398, \quad \tan \alpha = -4.2163702135578385, \quad \tan \alpha + \sin \alpha = -5.189378724378878
\][/tex]
