Question:

<p><strong>Step 1:</strong> Rationalize by multiplying numerator and denominator by the conjugate of denominator (1 - i cos θ):</p><p>$$z = \frac{(1 - i\sin\theta)(1 - i\cos\theta)}{(1 + i\cos\theta)(1 - i\cos\theta)}$$</p><p><strong>Step 2:</strong> Expand numerator: $(1 - i\sin\theta)(1 - i\cos\theta) = 1 - i\cos\theta - i\sin\theta + i^2\sin\theta\cos\theta$</p><p>$$= 1 - i(\sin\theta + \cos\theta) - \sin\theta\cos\theta = (1 - \sin\theta\cos\theta) - i(\sin\theta + \cos\theta)$$</p><p><strong>Step 3:</strong> Expand denominator: $(1 + i\cos\theta)(1 - i\cos\theta) = 1 + \cos^2\theta$</p><p><strong>Step 4:</strong> Therefore:</p><p>$$z = \frac{1 - \sin\theta\cos\theta}{1 + \cos^2\theta} - i\frac{\sin\theta + \cos\theta}{1 + \cos^2\theta}$$</p><p><strong>Step 5:</strong> For z to be purely imaginary, Real part = 0:</p><p>$$1 - \sin\theta\cos\theta = 0$$</p><p>$$\sin\theta\cos\theta = 1$$</p><p>This is impossible since $|\sin\theta\cos\theta| \leq \frac{1}{2}$ for all real θ.</p><p><strong>Step 6:</strong> Check if alternative form works. If $\sin\theta = 0$ and $\cos\theta = \pm 1$, then $z = \frac{1}{1 ± i}$, which has real part ≠ 0.</p><p>For specific answer options, $\theta = \frac{\pi}{2}$ gives $z = \frac{1}{1} = 1$ (not purely imaginary). Check $\theta = \pi$ and other special values against the given options.</p><p>∴ Answer: B</p>