\[ \sum_{m=1}^{6} \csc\left(\theta + (m-1)\frac{\pi}{4}\right) \csc\left(\theta + \frac{m\pi}{4}\right) = 4\sqrt{2} \] for \( 0 < \theta < \frac{\pi}{2} \). Here's a step-by-step explanation of the solution: 1. Simplify the Summation: The sum can be rewritten using the identity for the product of cosecants: \[ \csc A \csc B = \frac{1}{\sin A \sin B} \] Using the identity for the difference of sines: \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \] The sum becomes: \[ \sum_{m=1}^{6} \frac{\sin\left\{\left(\theta + \frac{m\pi}{4}\right) - \left(\theta + (m-1)\frac{\pi}{4}\right)\right\}}{\sin\left(\theta + \frac{m\pi}{4}\right) \sin\left(\theta + (m-1)\frac{\pi}{4}\right)} = 4 \] 2. Simplify the Argument of Sine: The argument simplifies to: \[ \left(\theta + \frac{m\pi}{4}\right) - \left(\theta + (m-1)\frac{\pi}{4}\right) = \frac{\pi}{4} \] So the sum becomes: \[ \sum_{m=1}^{6} \frac{\sin\left(\frac{\pi}{4}\right)}{\sin\left(\theta + \frac{m\pi}{4}\right) \sin\left(\theta + (m-1)\frac{\pi}{4}\right)} = 4 \] Since \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we have: \[ \sum_{m=1}^{6} \frac{\frac{\sqrt{2}}{2}}{\sin\left(\theta + \frac{m\pi}{4}\right) \sin\left(\theta + (m-1)\frac{\pi}{4}\right)} = 4 \] Simplifying further: \[ \sum_{m=1}^{6} \frac{1}{\sin\left(\theta + \frac{m\pi}{4}\right) \sin\left(\theta + (m-1)\frac{\pi}{4}\right)} = \frac{4}{\frac{\sqrt{2}}{2}} = 4\sqrt{2} \] 3. Use Trigonometric Identities: The sum can be expressed in terms of cotangent differences: \[ \sum_{m=1}^{6} \left\{\cot\left(\theta + (m-1)\frac{\pi}{4}\right) - \cot\left(\theta + \frac{m\pi}{4}\right)\right\} = 4 \] This is a telescoping series, where most terms cancel out, leaving: \[ \cot\theta - \cot\left(\theta + \frac{6\pi}{4}\right) = 4 \] Simplifying \(\cot\left(\theta + \frac{3\pi}{2}\right)\): \[ \cot\left(\theta + \frac{3\pi}{2}\right) = -\tan\theta \] So the equation becomes: \[ \cot\theta + \tan\theta = 4 \] 4. Solve for \(\theta\): Using the identity \(\cot\theta + \tan\theta = \frac{1}{\tan\theta} + \tan\theta\): \[ \frac{1}{\tan\theta} + \tan\theta = 4 \] Let \(x = \tan\theta\), then: \[ \frac{1}{x} + x = 4 \Rightarrow x^2 - 4x + 1 = 0 \] Solving the quadratic equation: \[ x = 2 \pm \sqrt{3} \] Therefore: \[ \theta = \arctan(2 \pm \sqrt{3}) \] The solutions within the interval \(0 < \theta < \frac{\pi}{2}\) are: \[ \theta = \frac{\pi}{12}, \frac{5\pi}{12} \] 5. Conclusion: The correct values of \(\theta\) are \(\frac{\pi}{12}\) and \(\frac{5\pi}{12}\), which correspond to options C and D. Therefore, the correct answer is: \[ \boxed{B} \]