1. **Problem Statement:** We are given the expression $\frac{1 - \tan^2 \theta}{\sec^2 \theta} = \cos 2\theta$ and asked to: a) State a non-permissible value of $\theta$. b) Prove the identity for all permissible values of $\theta$. 2. **Important Notes:** - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. - Non-permissible values occur where denominators are zero or expressions undefined. 3. **Part a) Non-permissible values:** - The denominator is $\sec^2 \theta = \frac{1}{\cos^2 \theta}$, so $\cos \theta \neq 0$. - Thus, $\theta \neq \frac{\pi}{2} + k\pi$, where $k$ is any integer. 4. **Part b) Prove the identity:** - Start with the left-hand side (LHS): $$\frac{1 - \tan^2 \theta}{\sec^2 \theta}$$ - Substitute $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$ and $\sec^2 \theta = \frac{1}{\cos^2 \theta}$: $$\frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{\frac{1}{\cos^2 \theta}}$$ - Simplify numerator: $$\frac{\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}}{\frac{1}{\cos^2 \theta}}$$ - Divide by fraction: $$\frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} \times \cancel{\frac{\cos^2 \theta}{1}}$$ - Cancel $\cos^2 \theta$: $$\cos^2 \theta - \sin^2 \theta$$ - Recall the double-angle identity for cosine: $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$ - Therefore, LHS $= \cos 2\theta$, which equals the right-hand side (RHS). 5. **Conclusion:** The identity holds for all $\theta$ where $\cos \theta \neq 0$.