1. **State the problem:** Find the value of $A$ such that the equation $\cos^2 x (1 + \cot^2 x) = A$ is an identity. 2. **Recall the Pythagorean identity:** We know that $1 + \cot^2 x = \csc^2 x$. 3. **Substitute into the equation:** $$\cos^2 x (1 + \cot^2 x) = \cos^2 x \cdot \csc^2 x$$ 4. **Rewrite $\csc^2 x$ in terms of sine:** $$\csc^2 x = \frac{1}{\sin^2 x}$$ 5. **Substitute and simplify:** $$\cos^2 x \cdot \frac{1}{\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$ 6. **Conclusion:** The expression simplifies to $\cot^2 x$, so the value of $A$ must be $\cot^2 x$. **Final answer:** $A = \cot^2 x$ --- 1. **State the problem:** Prove the identity: $$\frac{1}{1 + \sin \theta} = \frac{\sec \theta - \sin \theta \sec \theta}{\cos \theta}$$ 2. **Rewrite the right side:** Recall $\sec \theta = \frac{1}{\cos \theta}$, so: $$\frac{\sec \theta - \sin \theta \sec \theta}{\cos \theta} = \frac{\frac{1}{\cos \theta} - \sin \theta \cdot \frac{1}{\cos \theta}}{\cos \theta} = \frac{\frac{1 - \sin \theta}{\cos \theta}}{\cos \theta}$$ 3. **Simplify the complex fraction:** $$= \frac{1 - \sin \theta}{\cos \theta} \cdot \frac{1}{\cos \theta} = \frac{1 - \sin \theta}{\cos^2 \theta}$$ 4. **Rewrite the left side:** Multiply numerator and denominator by $1 - \sin \theta$: $$\frac{1}{1 + \sin \theta} \cdot \frac{1 - \sin \theta}{1 - \sin \theta} = \frac{1 - \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)}$$ 5. **Simplify the denominator:** $$(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta$$ 6. **So the left side becomes:** $$\frac{1 - \sin \theta}{\cos^2 \theta}$$ 7. **Both sides are equal:** $$\frac{1}{1 + \sin \theta} = \frac{1 - \sin \theta}{\cos^2 \theta} = \frac{\sec \theta - \sin \theta \sec \theta}{\cos \theta}$$ **Non-permissible values:** - $\cos \theta = 0$ (denominator zero in $\sec \theta$ and right side) - $1 + \sin \theta = 0 \Rightarrow \sin \theta = -1$ (denominator zero in left side) Hence, $\theta \neq \frac{\pi}{2} + k\pi$ and $\theta \neq \frac{3\pi}{2} + 2k\pi$ for integers $k$.