1. **State the problem:** Prove that $$\sec^2\theta - 2\tan^2\theta + 2\tan\theta = 0$$ and find the values of $$\theta$$. 2. **Recall identities:** We know that $$\sec^2\theta = 1 + \tan^2\theta$$. 3. **Substitute:** Replace $$\sec^2\theta$$ in the equation: $$1 + \tan^2\theta - 2\tan^2\theta + 2\tan\theta = 0$$ 4. **Simplify:** $$1 - \tan^2\theta + 2\tan\theta = 0$$ 5. **Rewrite as a quadratic in $$x = \tan\theta$$:** $$-x^2 + 2x + 1 = 0$$ Multiply both sides by $$-1$$ to get: $$x^2 - 2x - 1 = 0$$ 6. **Solve quadratic:** Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1$$, $$b=-2$$, $$c=-1$$: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 7. **Find $$\theta$$:** $$\tan\theta = 1 + \sqrt{2}$$ or $$\tan\theta = 1 - \sqrt{2}$$ Therefore, $$\theta = \arctan(1 + \sqrt{2}) + k\pi$$ or $$\theta = \arctan(1 - \sqrt{2}) + k\pi$$ for any integer $$k$$. **Final answer:** The equation is proven and the solutions for $$\theta$$ are $$\theta = \arctan(1 + \sqrt{2}) + k\pi$$ or $$\theta = \arctan(1 - \sqrt{2}) + k\pi$$ where $$k$$ is any integer.