If sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ−cos⁡θ1+cot2⁡θ−2tan⁡θcot⁡θ=−1,θ∈[0,2π],then

sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ=sin2⁡θ+cos2⁡θ+sin⁡θcos⁡θθ≠π4,5π4=1+sin⁡θcos⁡θand cos⁡θ1+cot2⁡θ=cos⁡θ|cosec⁡θ|=sin⁡θcos⁡θ∀θ∈(0,π)and −2tan⁡θcot⁡θ=−2θ≠π2hence LHS=RHSbut θ≠π4,5π4,π2Hence θ∈(0,π)~π4,π2

If sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ−cos⁡θ1+cot2⁡θ−2tan⁡θcot⁡θ=−1,θ∈[0,2π],then

sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ=sin2⁡θ+cos2⁡θ+sin⁡θcos⁡θθ≠π4,5π4=1+sin⁡θcos⁡θand cos⁡θ1+cot2⁡θ=cos⁡θ|cosec⁡θ|=sin⁡θcos⁡θ∀θ∈(0,π)and −2tan⁡θcot⁡θ=−2θ≠π2hence LHS=RHSbut θ≠π4,5π4,π2Hence θ∈(0,π)~π4,π2

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If $\frac{\sin^{3} θ - \cos^{3} θ}{\sin θ - \cos θ} - \frac{\cos θ}{\sqrt{(1 + \cot^{2} θ)}} - 2 \tan θcot θ = - 1, θ \in [0, 2 π]$ ,then

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$θ \in (0, \frac{π}{2}) - \{\frac{π}{4}\}$

$θ \in (\frac{π}{2}, π) - \{\frac{3 π}{4}\}$

$θ \in (π, \frac{3 π}{2}) - \{\frac{5 π}{4}\}$

$θ \in (0, π) - \{\frac{π}{4}, \frac{π}{2}\}$

answer is D.

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Detailed Solution

$\frac{\sin^{3} θ - \cos^{3} θ}{\sin θ - \cos θ} = \sin^{2} θ + \cos^{2} θ + \sin θcos θ$
$\begin{array}{l} θ \neq \frac{π}{4}, \frac{5 π}{4} \\ = 1 + \sin θcos θ \end{array}$
$\begin{array}{l} and \frac{\cos θ}{\sqrt{(1 + \cot^{2} θ)}} = \frac{\cos θ}{| cosec θ |} = \sin θcos θ \forall θ \in (0, π) \\ and - 2 \tan θcot θ = - 2 θ \neq \frac{π}{2} \end{array}$