1. **State the problem:** We are given that $\sin \theta = \frac{\sqrt{2}}{2}$ and need to find $\tan \theta$. 2. **Recall the definitions and formulas:** - $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ 3. **Find $\cos \theta$ using the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ Substitute $\sin \theta = \frac{\sqrt{2}}{2}$: $$\left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2 \theta = 1$$ $$\frac{2}{4} + \cos^2 \theta = 1$$ $$\frac{1}{2} + \cos^2 \theta = 1$$ $$\cos^2 \theta = 1 - \frac{1}{2} = \frac{1}{2}$$ 4. **Take the square root to find $\cos \theta$:** $$\cos \theta = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$ 5. **Calculate $\tan \theta$:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{2}}{2}}{\pm \frac{\sqrt{2}}{2}}$$ 6. **Simplify the fraction:** $$\tan \theta = \frac{\cancel{\frac{\sqrt{2}}{2}}}{\pm \cancel{\frac{\sqrt{2}}{2}}} = \pm 1$$ **Final answer:** $$\tan \theta = \pm 1$$ The sign depends on the quadrant where $\theta$ lies, but the magnitude is 1.