1. **Problem statement:** (a) Differentiate $y = \frac{\cos x}{\sin x}$ and show that $\frac{dy}{dx} = -\csc^2 x$. (b) Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc^2 2x \, dx$ and show it equals $\frac{\sqrt{3}}{6}$. 2. **Part (a) Differentiation:** We start with the quotient rule for differentiation: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $u = \cos x$ and $v = \sin x$. 3. Calculate derivatives of numerator and denominator: $$\frac{du}{dx} = -\sin x, \quad \frac{dv}{dx} = \cos x$$ 4. Apply the quotient rule: $$\frac{dy}{dx} = \frac{\sin x (-\sin x) - \cos x (\cos x)}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$$ 5. Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$: $$\frac{dy}{dx} = \frac{-1}{\sin^2 x} = -\csc^2 x$$ 6. **Part (b) Integration:** Evaluate: $$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc^2 2x \, dx$$ 7. Use substitution $u = 2x$, so $du = 2 dx$ or $dx = \frac{du}{2}$. Change limits: When $x = \frac{\pi}{6}$, $u = \frac{\pi}{3}$. When $x = \frac{\pi}{4}$, $u = \frac{\pi}{2}$. 8. Rewrite the integral: $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \csc^2 u \cdot \frac{1}{2} du = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \csc^2 u \, du$$ 9. Recall the integral formula: $$\int \csc^2 u \, du = -\cot u + C$$ 10. Evaluate the definite integral: $$\frac{1}{2} \left[-\cot u \right]_{\frac{\pi}{3}}^{\frac{\pi}{2}} = \frac{1}{2} \left(-\cot \frac{\pi}{2} + \cot \frac{\pi}{3} \right)$$ 11. Calculate values: $$\cot \frac{\pi}{2} = 0, \quad \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$$ 12. Substitute back: $$\frac{1}{2} (0 + \frac{1}{\sqrt{3}}) = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}$$ **Final answers:** (a) $\frac{dy}{dx} = -\csc^2 x$ (b) $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc^2 2x \, dx = \frac{\sqrt{3}}{6}$