1. We are asked to calculate the tangent and cotangent of angle $\alpha$ given that $\sin \alpha = \frac{1}{6}$ and that $\alpha$ lies in the first quadrant. 2. Recall the definitions: - $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ - $\cot \alpha = \frac{1}{\tan \alpha} = \frac{\cos \alpha}{\sin \alpha}$ 3. Since $\sin \alpha = \frac{1}{6}$, we can find $\cos \alpha$ using the Pythagorean identity: $$\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{1}{6}\right)^2} = \sqrt{1 - \frac{1}{36}} = \sqrt{\frac{35}{36}} = \frac{\sqrt{35}}{6}$$ 4. Now calculate $\tan \alpha$: $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{1}{6}}{\frac{\sqrt{35}}{6}} = \frac{1}{6} \times \frac{6}{\sqrt{35}} = \frac{1}{\sqrt{35}}$$ 5. Simplify $\tan \alpha$ by rationalizing the denominator: $$\tan \alpha = \frac{1}{\sqrt{35}} \times \frac{\sqrt{35}}{\sqrt{35}} = \frac{\sqrt{35}}{35}$$ 6. Calculate $\cot \alpha$: $$\cot \alpha = \frac{1}{\tan \alpha} = \sqrt{35}$$ Final answers: $$\tan \alpha = \frac{\sqrt{35}}{35}$$ $$\cot \alpha = \sqrt{35}$$