1. **Problem:** Determine whether the equation $\frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta + \sin \theta} = \cos \theta - \sin \theta$ is an identity or a conditional equation. Step 1: Use the Pythagorean identity $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$. Rewrite numerator: $$\cos^2 \theta - \sin^2 \theta = \cos 2\theta$$ Step 2: The left side becomes $$\frac{\cos 2\theta}{\cos \theta + \sin \theta}$$ Step 3: The right side is $\cos \theta - \sin \theta$. Step 4: Check if both sides are equal generally (for all $\theta$), or find exceptions. Multiply both sides by $(\cos \theta + \sin \theta)$: $$\cos 2\theta = (\cos \theta - \sin \theta)(\cos \theta + \sin \theta)$$ Step 5: Expand right side: $$ (\cos \theta)^2 - (\sin \theta)^2 = \cos^2 \theta - \sin^2 \theta$$ Step 6: Using Pythagorean identity, the right side is $\cos^2 \theta - \sin^2 \theta$, which equals $\cos 2\theta$. Step 7: Left side equals right side as long as $\cos \theta + \sin \theta \neq 0$ (denominator not zero). Step 8: Find values where denominator zero: $$\cos \theta + \sin \theta = 0 \implies \sin \theta = -\cos \theta \implies \tan \theta = -1$$ These values are $\theta = -\frac{\pi}{4} + k\pi$, where $k$ is any integer. **Conclusion:** The equation is an identity for all $\theta$ except where denominator is zero. 2. **Problem:** Determine whether $\frac{x+2}{x+3} = 1 - \frac{1}{x^2 + 2x - 3}$ is identity or conditional. Step 1: Factor denominator on right: $$x^2 + 2x - 3 = (x+3)(x-1)$$ Step 2: Write right side with common denominator: $$1 - \frac{1}{(x+3)(x-1)} = \frac{(x+3)(x-1) - 1}{(x+3)(x-1)}$$ Step 3: Expand numerator: $$ (x+3)(x-1) -1 = (x^2 + 2x -3) -1 = x^2 + 2x -4$$ Step 4: Left side denominator is $(x+3)$, multiply both sides by $(x+3)(x-1)$: $$ (x+2)(x-1) = x^2 + 2x -4$$ Step 5: Expand left: $$ x^2 + 2x - x - 2 = x^2 + x - 2$$ Step 6: Compare both sides: $$x^2 + x - 2 = x^2 + 2x -4$$ Step 7: Simplify: $$x^2 + x - 2 - (x^2 + 2x -4) = 0 \implies x - 2x + (-2 + 4) = 0 \implies -x + 2 = 0$$ Step 8: Solve for $x$: $$-x + 2 = 0 \implies x = 2$$ Step 9: Check domain restrictions: Denominator $x+3 \neq 0 \Rightarrow x \neq -3$. Denominator on right $(x+3)(x-1) \neq 0 \Rightarrow x \neq -3, 1$. Step 10: Test $x=2$ in original equation: Left: $\frac{2+2}{2+3} = \frac{4}{5} = 0.8$ Right: $$1 - \frac{1}{(2)^2 + 2(2) - 3} = 1 - \frac{1}{4 + 4 -3} = 1 - \frac{1}{5} = \frac{4}{5} = 0.8$$ Valid. Step 11: Check if identity by testing other values like $x=0$: Left: $\frac{0+2}{0+3} = \frac{2}{3} \approx 0.666...$ Right: $$1 - \frac{1}{0 + 0 - 3} = 1 - \frac{1}{-3} = 1 + \frac{1}{3} = \frac{4}{3} \neq 0.666...$$ So not an identity, but conditional equation true only when $x=2$. 3. **Problem:** Check if $\sin x \cos x = 1$ is identity or conditional. Step 1: Note $|\sin x| \leq 1$ and $|\cos x| \leq 1$ so $|\sin x \cos x| \leq 1$. Step 2: Maximum magnitude of $\sin x \cos x$ is $\frac{1}{2}$ (using double angle identity): $$\sin x \cos x = \frac{1}{2} \sin 2x$$ The max of $|\sin 2x|$ is 1. Step 3: So max $|\sin x \cos x| = \frac{1}{2} < 1$. Step 4: Therefore, $\sin x \cos x = 1$ has no solution in real numbers (no $x$ satisfies the equality). **Conclusion:** Conditional equation with no real solution. 4. **Problem:** Determine if $2 - \sin^2 x = \sec x + \cos x$ is identity or conditional. Step 1: Recall $\sec x = \frac{1}{\cos x}$. Step 2: Rewrite left side: $$ 2 - \sin^2 x = 2 - (1 - \cos^2 x) = 2 - 1 + \cos^2 x = 1 + \cos^2 x$$ Step 3: Right side: $$ \frac{1}{\cos x} + \cos x$$ Step 4: Multiply both sides by $\cos x$ (domain: $\cos x \neq 0$): $$ \cos x (1 + \cos^2 x) = 1 + \cos^2 x$$ Step 5: Left: $$ \cos x + \cos^3 x$$ Step 6: Right is: $$1 + \cos^2 x$$ Step 7: Rearrange: $$ \cos x + \cos^3 x - 1 - \cos^2 x = 0$$ Step 8: Let $y = \cos x$: $$ y + y^3 - 1 - y^2 = 0 \implies y^3 - y^2 + y -1 = 0$$ Step 9: Factor polynomial: $$ (y^3 - y^2) + (y - 1) = y^2(y-1) + (y -1) = (y -1)(y^2 + 1) = 0$$ Step 10: Solve: $$ y - 1 = 0 \implies y = 1$$ or $$ y^2 +1 = 0 \implies y^2 = -1$$ (impossible for real cosines) Step 11: Thus, $\cos x = 1$ is only solution. Step 12: Check domain restrictions: $\cos x \neq 0$ to avoid $\sec x$ undefined. Step 13: At $\cos x =1$, original equation: Left: $2 - \sin^2 x = 2 - 0 = 2$ Right: $\sec x + \cos x = 1 + 1 = 2$ Valid. **Conclusion:** Conditional equation true only when $\cos x = 1$, i.e., $x = 2k\pi$. **Final answers:** 1. Identity (valid except where denominator zero: $\theta = -\frac{\pi}{4} + k\pi$) 2. Conditional, true only at $x=2$ (domain excludes $x = -3, 1$). 3. Conditional, no real solution. 4. Conditional, true only when $\cos x = 1$ ($x=2k\pi$).