1. **State the problem:** We are given two identities involving trigonometric functions: $$\sin^2 \alpha = \frac{1}{1 + \tan^2 \alpha}$$ and $$\tan^2 \alpha = \frac{1}{\sin^2 \alpha} - 1$$ We need to verify these identities and solve for the relationships between $\sin^2 \alpha$ and $\tan^2 \alpha$. 2. **Recall the Pythagorean identity:** $$1 + \tan^2 \alpha = \sec^2 \alpha$$ and $$\sin^2 \alpha + \cos^2 \alpha = 1$$ 3. **Start with the first identity:** $$\sin^2 \alpha = \frac{1}{1 + \tan^2 \alpha}$$ Using the Pythagorean identity, substitute $1 + \tan^2 \alpha = \sec^2 \alpha$: $$\sin^2 \alpha = \frac{1}{\sec^2 \alpha}$$ Since $\sec \alpha = \frac{1}{\cos \alpha}$, then $\sec^2 \alpha = \frac{1}{\cos^2 \alpha}$, so: $$\sin^2 \alpha = \cos^2 \alpha$$ 4. **Check if $\sin^2 \alpha = \cos^2 \alpha$ is true:** This equality holds when $\sin^2 \alpha - \cos^2 \alpha = 0$, or equivalently: $$\cos 2\alpha = 0$$ which means: $$2\alpha = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ 5. **Now verify the second identity:** $$\tan^2 \alpha = \frac{1}{\sin^2 \alpha} - 1$$ Rewrite the right side: $$\frac{1}{\sin^2 \alpha} - 1 = \frac{1 - \sin^2 \alpha}{\sin^2 \alpha} = \frac{\cos^2 \alpha}{\sin^2 \alpha} = \cot^2 \alpha$$ 6. **Compare with the left side:** $$\tan^2 \alpha = \cot^2 \alpha$$ This implies: $$\tan^2 \alpha = \frac{1}{\tan^2 \alpha}$$ Multiply both sides by $\tan^2 \alpha$: $$\tan^4 \alpha = 1$$ Take the square root: $$\tan^2 \alpha = 1$$ 7. **Summary of solutions:** - From step 4, $\sin^2 \alpha = \cos^2 \alpha$ implies $\alpha = \frac{\pi}{4} + \frac{k\pi}{2}$. - From step 6, $\tan^2 \alpha = 1$ implies $\tan \alpha = \pm 1$, which matches the same angles. **Final answer:** $$\boxed{\alpha = \frac{\pi}{4} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}}$$