If sin⁡3θ−cos⁡3θsin⁡θ−cos⁡θ−cos⁡θ(1+cot⁡2θ)−2tan⁡θcot⁡θ=−1\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{(1+\cot^2 \theta)}} - 2 \tan \theta \cot \theta = -1…

Problem

If sin⁡3θ−cos⁡3θsin⁡θ−cos⁡θ−cos⁡θ(1+cot⁡2θ)−2tan⁡θcot⁡θ=−1\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{(1+\cot^2 \theta)}} - 2 \tan \theta \cot \theta = -1, θ∈[0,2π]\theta \in [0, 2\pi], then a. θ∈(0,π2)−{π4}\theta \in \left(0, \frac{\pi}{2}\right) - \left\{\frac{\pi}{4}\right\} b. θ∈(π2,π)−{3π4}\theta \in \left(\frac{\pi}{2}, \pi\right) - \left\{\frac{3\pi}{4}\right\} c. θ∈(π,3π2)−{5π4}\theta \in \left(\pi, \frac{3\pi}{2}\right) - \left\{\frac{5\pi}{4}\right\} d. θ∈(0,π)−{π4,π2}\theta \in (0, \pi) - \left\{\frac{\pi}{4}, \frac{\pi}{2}\right\}

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