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

∵sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ=(sin⁡θ−cos⁡θ)sin2⁡θ+cos2⁡θ+sin⁡θcos⁡θ(sin⁡θ−cos⁡θ)=1+sin⁡θcos⁡θ, when θ≠π4,5π4 and −cos⁡θ1+cot2⁡θ=−cos⁡θ|cosec⁡θ|=−sin⁡θcos⁡θ, when θ∈(0,π) and −2tan⁡θcot⁡θ=−2, when θ≠0,π2 Now, sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ−cos⁡θ1+cot2⁡θ−2tan⁡θcot⁡θ=−1 When, θ∈(0,π) and θ≠0,π4,5π4,π2∴ 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 θ belongs to

∵sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ=(sin⁡θ−cos⁡θ)sin2⁡θ+cos2⁡θ+sin⁡θcos⁡θ(sin⁡θ−cos⁡θ)=1+sin⁡θcos⁡θ, when θ≠π4,5π4 and −cos⁡θ1+cot2⁡θ=−cos⁡θ|cosec⁡θ|=−sin⁡θcos⁡θ, when θ∈(0,π) and −2tan⁡θcot⁡θ=−2, when θ≠0,π2 Now, sin3⁡θ−cos3⁡θsin⁡θ−cos⁡θ−cos⁡θ1+cot2⁡θ−2tan⁡θcot⁡θ=−1 When, θ∈(0,π) and θ≠0,π4,5π4,π2∴ θ∈(0,π)−π4,π2

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

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

$(0, π)$

None of the above

$[0, π] - \{\frac{π}{4}, \frac{π}{2}\}$

answer is C.

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Detailed Solution

$\begin{array}{l} ∵ \frac{\sin^{3} θ - \cos^{3} θ}{\sin θ - \cos θ} \\ = \frac{(\sin θ - \cos θ) (\sin^{2} θ + \cos^{2} θ + \sin θ \cos θ)}{(\sin θ - \cos θ)} \\ = 1 + \sin θ \cos θ, when θ \neq \frac{π}{4}, \frac{5 π}{4} \end{array}$

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