1. The problem is to simplify and solve the expression $\sec \theta - 1 = 1 - \tan^2 \theta$ for $\theta$. 2. Recall the Pythagorean identity: $$1 + \tan^2 \theta = \sec^2 \theta$$ 3. Rearranging the identity gives: $$\tan^2 \theta = \sec^2 \theta - 1$$ 4. Substitute $\tan^2 \theta$ in the original equation: $$\sec \theta - 1 = 1 - (\sec^2 \theta - 1)$$ 5. Simplify the right side: $$\sec \theta - 1 = 1 - \sec^2 \theta + 1 = 2 - \sec^2 \theta$$ 6. Bring all terms to one side: $$\sec \theta - 1 - 2 + \sec^2 \theta = 0$$ 7. Simplify: $$\sec^2 \theta + \sec \theta - 3 = 0$$ 8. Let $x = \sec \theta$, then the quadratic equation is: $$x^2 + x - 3 = 0$$ 9. Solve the quadratic using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2}$$ 10. So, $$x_1 = \frac{-1 + \sqrt{13}}{2}, \quad x_2 = \frac{-1 - \sqrt{13}}{2}$$ 11. Since $\sec \theta = \frac{1}{\cos \theta}$, and $\sec \theta$ must be real and defined, we consider values of $x$ where $|x| \geq 1$. 12. Evaluate approximate values: $$x_1 \approx \frac{-1 + 3.606}{2} = 1.303 > 1$$ $$x_2 \approx \frac{-1 - 3.606}{2} = -2.303 < -1$$ 13. Both values are valid for $\sec \theta$. 14. Therefore, $$\cos \theta = \frac{1}{x}$$ 15. Final solutions: $$\cos \theta = \frac{2}{-1 + \sqrt{13}} \quad \text{or} \quad \cos \theta = \frac{2}{-1 - \sqrt{13}}$$ These correspond to the angles $\theta$ where the original equation holds true.