1. **Rewrite** $\sqrt{x^2 + 36}$ using the substitution $x = 6 \tan \theta$, where $0 < \theta < \frac{\pi}{2}$. 2. Substitute $x = 6 \tan \theta$ into the expression: $$\sqrt{(6 \tan \theta)^2 + 36} = \sqrt{36 \tan^2 \theta + 36}$$ 3. Factor out 36 inside the square root: $$\sqrt{36 (\tan^2 \theta + 1)}$$ 4. Use the Pythagorean identity $1 + \tan^2 \theta = \sec^2 \theta$: $$\sqrt{36 \sec^2 \theta}$$ 5. Simplify the square root: $$6 \sec \theta$$ **Answer:** $\sqrt{x^2 + 36} = 6 \sec \theta$. --- 1. **Simplify** $\ln |\cos \theta| - \ln |\sin \theta|$. 2. Use the logarithm subtraction rule: $$\ln |\cos \theta| - \ln |\sin \theta| = \ln \left| \frac{\cos \theta}{\sin \theta} \right|$$ 3. Recognize that $\frac{\cos \theta}{\sin \theta} = \cot \theta$: $$\ln |\cot \theta|$$ **Answer:** $\ln |\cos \theta| - \ln |\sin \theta| = \ln |\cot \theta|$. --- 1. **Simplify** $$y = \frac{\cos x}{1 - \tan x} + \frac{\sin x \cos x}{\sin x - \cos x}$$ 2. Simplify the first fraction by multiplying numerator and denominator by $\cos x$: $$\frac{\cos x}{1 - \tan x} = \frac{\cos x}{1 - \frac{\sin x}{\cos x}} = \frac{\cos x}{\frac{\cos x - \sin x}{\cos x}} = \frac{\cos x \cancel{\cos x}}{\cos x - \sin x} = \frac{\cos^2 x}{\cos x - \sin x}$$ 3. The second fraction is: $$\frac{\sin x \cos x}{\sin x - \cos x} = \frac{\sin x \cos x}{-(\cos x - \sin x)} = - \frac{\sin x \cos x}{\cos x - \sin x}$$ 4. Combine the two fractions over common denominator $\cos x - \sin x$: $$y = \frac{\cos^2 x}{\cos x - \sin x} - \frac{\sin x \cos x}{\cos x - \sin x} = \frac{\cos^2 x - \sin x \cos x}{\cos x - \sin x}$$ 5. Factor numerator: $$\cos x (\cos x - \sin x)$$ 6. Cancel common factor $\cos x - \sin x$: $$\frac{\cos x \cancel{(\cos x - \sin x)}}{\cancel{\cos x - \sin x}} = \cos x$$ **Answer:** $y = \cos x$. --- 1. **Simplify** $$y = \sin x + \frac{\cot^2 x}{\csc x}$$ 2. Express $\cot x$ and $\csc x$ in terms of sine and cosine: $$\cot x = \frac{\cos x}{\sin x}, \quad \csc x = \frac{1}{\sin x}$$ 3. Substitute: $$y = \sin x + \frac{\left( \frac{\cos x}{\sin x} \right)^2}{\frac{1}{\sin x}} = \sin x + \frac{\frac{\cos^2 x}{\sin^2 x}}{\frac{1}{\sin x}}$$ 4. Simplify the complex fraction: $$\frac{\frac{\cos^2 x}{\sin^2 x}}{\frac{1}{\sin x}} = \frac{\cos^2 x}{\sin^2 x} \times \sin x = \frac{\cos^2 x}{\sin x}$$ 5. So, $$y = \sin x + \frac{\cos^2 x}{\sin x} = \frac{\sin^2 x}{\sin x} + \frac{\cos^2 x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x}$$ 6. Use Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $$y = \frac{1}{\sin x} = \csc x$$ **Answer:** $y = \csc x$. --- 1. **Find** $\sin^2 25^\circ + \sin^2 65^\circ$. 2. Use the identity $\sin^2 A + \sin^2 B = 1 - \cos(2A) - \cos(2B)$ is complicated, so instead use the complementary angle: Since $65^\circ = 90^\circ - 25^\circ$, $\sin 65^\circ = \cos 25^\circ$. 3. So, $$\sin^2 25^\circ + \sin^2 65^\circ = \sin^2 25^\circ + \cos^2 25^\circ = 1$$ **Answer:** $\sin^2 25^\circ + \sin^2 65^\circ = 1$. --- 1. **Find** $\cos^2 20^\circ + \cos^2 52^\circ + \cos^2 38^\circ + \cos^2 70^\circ$. 2. Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ for each term: $$\cos^2 20^\circ = \frac{1 + \cos 40^\circ}{2}$$ $$\cos^2 52^\circ = \frac{1 + \cos 104^\circ}{2}$$ $$\cos^2 38^\circ = \frac{1 + \cos 76^\circ}{2}$$ $$\cos^2 70^\circ = \frac{1 + \cos 140^\circ}{2}$$ 3. Sum all: $$\sum = \frac{1 + \cos 40^\circ}{2} + \frac{1 + \cos 104^\circ}{2} + \frac{1 + \cos 76^\circ}{2} + \frac{1 + \cos 140^\circ}{2} = 2 + \frac{\cos 40^\circ + \cos 104^\circ + \cos 76^\circ + \cos 140^\circ}{2}$$ 4. Use cosine sum properties or approximate values: $\cos 40^\circ \approx 0.7660$, $\cos 104^\circ \approx -0.2419$, $\cos 76^\circ \approx 0.2419$, $\cos 140^\circ \approx -0.7660$. 5. Sum cosines: $$0.7660 - 0.2419 + 0.2419 - 0.7660 = 0$$ 6. So, $$\sum = 2 + \frac{0}{2} = 2$$ **Answer:** $\cos^2 20^\circ + \cos^2 52^\circ + \cos^2 38^\circ + \cos^2 70^\circ = 2$.