Question 9: Prove that cotθ + tanθ = 2cosec2θ. Hence show that ∫(π/6)^(π/3) cosec2θ dθ = (1/2)ln3

here's the solution to Question 9:

Part (i): Prove that $\cot\theta + \tan\theta = 2\operatorname{cosec}2\theta$.

Step 1.1: Start with the left-hand side (LHS) and express $\cot\theta$ and $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$. $\text{LHS} = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}$

Step 1.2: Combine the fractions by finding a common denominator. $\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta}$

Step 1.3: Use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. $\frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$

Step 1.4: Use the double angle identity for sine, $\sin2\theta = 2\sin\theta\cos\theta$, which implies $\sin\theta\cos\theta = \frac{1}{2}\sin2\theta$. $\frac{1}{\sin\theta\cos\theta} = \frac{1}{\frac{1}{2}\sin2\theta} = \frac{2}{\sin2\theta}$

Step 1.5: Express the result in terms of $\operatorname{cosec}2\theta$, since $\operatorname{cosec}x = \frac{1}{\sin x}$. $\frac{2}{\sin2\theta} = 2\operatorname{cosec}2\theta = \text{RHS}$ Thus, $\cot\theta + \tan\theta = \boxed{2\operatorname{cosec}2\theta}$ is proven.

Part (ii): Hence show that $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \operatorname{cosec}2\theta\,d\theta = \frac{1}{2}\ln3$.

Step 2.1: Use the result from Part (i) to rewrite the integrand. From Part (i), we have $2\operatorname{cosec}2\theta = \cot\theta + \tan\theta$, so $\operatorname{cosec}2\theta = \frac{1}{2}(\cot\theta + \tan\theta)$. $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \operatorname{cosec}2\theta\,d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{2}(\cot\theta + \tan\theta)\,d\theta$

Step 2.2: Integrate the terms. Recall that $\int \cot\theta\,d\theta = \ln|\sin\theta|$ and $\int \tan\theta\,d\theta = -\ln|\cos\theta|$. $\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\cot\theta + \tan\theta)\,d\theta = \frac{1}{2} \left[ \ln|\sin\theta| - \ln|\cos\theta| \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

Step 2.3: Use the logarithm property $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$. $\frac{1}{2} \left[ \ln\left|\frac{\sin\theta}{\cos\theta}\right| \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{2} \left[ \ln|\tan\theta| \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

Step 2.4: Apply the limits of integration. $\frac{1}{2} \left( \ln\left|\tan\left(\frac{\pi}{3}\right)\right| - \ln\left|\tan\left(\frac{\pi}{6}\right)\right| \right)$

Step 2.5: Evaluate the tangent values: $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. $\frac{1}{2} \left( \ln(\sqrt{3}) - \ln\left(\frac{1}{\sqrt{3}}\right) \right)$

Step 2.6: Simplify the logarithmic expression using $\ln\left(\frac{1}{a}\right) = -\ln a$ and $\ln(a^b) = b\ln a$. $\frac{1}{2} \left( \ln(3^{\frac{1}{2}}) - \ln(3^{-\frac{1}{2}}) \right) = \frac{1}{2} \left( \frac{1}{2}\ln3 - \left(-\frac{1}{2}\ln3\right) \right)$ $= \frac{1}{2} \left( \frac{1}{2}\ln3 + \frac{1}{2}\ln3 \right) = \frac{1}{2} (\ln3) = \boxed{\frac{1}{2}\ln3}$ The result is shown.