Question:

<p><strong>Step 1:</strong> Write out the given definitions clearly:</p><p>u_n = sin(nθ)sec^n(θ), v_n = cos(nθ)sec^n(θ)</p><p>u_{n-1} = sin((n-1)θ)sec^{n-1}(θ), v_{n-1} = cos((n-1)θ)sec^{n-1}(θ)</p><p><strong>Step 2:</strong> Simplify the first term: $\frac{v_n - v_{n-1}}{u_{n-1}}$</p><p>$v_n - v_{n-1} = \cos(n\theta)\sec^n(\theta) - \cos((n-1)\theta)\sec^{n-1}(\theta)$</p><p>$= \sec^{n-1}(\theta)[\cos(n\theta)\sec(\theta) - \cos((n-1)\theta)]$</p><p>$= \sec^{n-1}(\theta)[\frac{\cos(n\theta)}{\cos\theta} - \cos((n-1)\theta)]$</p><p><strong>Step 3:</strong> Apply the identity: $\cos(n\theta) - \cos((n-1)\theta)\cos\theta = -\sin((n-1)\theta)\sin\theta$</p><p>Using: $\cos(n\theta) = \cos((n-1)\theta + \theta) = \cos((n-1)\theta)\cos\theta - \sin((n-1)\theta)\sin\theta$</p><p>Therefore: $\cos(n\theta) - \cos((n-1)\theta)\cos\theta = -\sin((n-1)\theta)\sin\theta$</p><p>Thus: $v_n - v_{n-1} = -\sec^{n-1}(\theta)\sin((n-1)\theta)\sin\theta$</p><p><strong>Step 4:</strong> Compute the first fraction:</p><p>$\frac{v_n - v_{n-1}}{u_{n-1}} = \frac{-\sec^{n-1}(\theta)\sin((n-1)\theta)\sin\theta}{\sin((n-1)\theta)\sec^{n-1}(\theta)} = -\sin\theta/\cos\theta = -\cot\theta$</p><p><strong>Step 5:</strong> Simplify the second term: $\frac{1}{n}\frac{u_n}{v_n}$</p><p>$\frac{u_n}{v_n} = \frac{\sin(n\theta)\sec^n(\theta)}{\cos(n\theta)\sec^n(\theta)} = \frac{\sin(n\theta)}{\cos(n\theta)} = \tan(n\theta)$</p><p>Therefore: $\frac{1}{n}\frac{u_n}{v_n} = \frac{1}{n}\tan(n\theta)$</p><p><strong>Step 6:</strong> Add the two parts:</p><p>$\frac{v_n - v_{n-1}}{u_{n-1}} + \frac{1}{n}\frac{u_n}{v_n} = -\cot\theta + \frac{1}{n}\tan(n\theta)$</p><p><strong>∴ Answer: A</strong></p>