If sin3θ−cos3θsinθ−cosθ−cosθ1+cot2θ−2tanθcotθ=−1,θ∈[0,2π], then θ belong to

∵sin3θ−cos3θsinθ−cosθ=(sinθ−cosθ)(sin2θ+cos2θ+sinθcosθ)(sinθ−cosθ)=1+sinθcosθ,when θ≠π4,5π4 and −cosθ1+cot2θ=−cosθ|cosecθ|=−sinθcosθ,when θ∈(0,π) and −2tanθcotθ=−2,whenθ≠0,π2 Now, sin3θ−cos3θsinθ−cosθ−cosθ1+cot2θ−2tanθcotθ=−1When, θ∈(0,π)and θ≠0,π4,5π4,π2 ∴θ∈(0,π)−{π4,π2}

If sin3θ−cos3θsinθ−cosθ−cosθ1+cot2θ−2tanθcotθ=−1,θ∈[0,2π], then θ belong to

∵sin3θ−cos3θsinθ−cosθ=(sinθ−cosθ)(sin2θ+cos2θ+sinθcosθ)(sinθ−cosθ)=1+sinθcosθ,when θ≠π4,5π4 and −cosθ1+cot2θ=−cosθ|cosecθ|=−sinθcosθ,when θ∈(0,π) and −2tanθcotθ=−2,whenθ≠0,π2 Now, sin3θ−cos3θsinθ−cosθ−cosθ1+cot2θ−2tanθcotθ=−1When, 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 $θ$ belong to

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$(0, π) - {\frac{π}{4}, \frac{π}{2}}$

$(0, π)$

$[0, π] - [\frac{π}{4}, \frac{π}{2}]$

$[\frac{π}{4}, \frac{π}{2}]$

answer is C.

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Detailed Solution

$∵ \frac{\sin^{3} θ - \cos^{3} θ}{\sin θ - \cos θ} = \frac{(\sin θ - \cos θ) (\sin^{2} θ + \cos^{2} θ + \sin θ \cos θ)}{(\sin θ - \cos θ)} = 1 + \sin θ \cos θ, w h e n θ \neq \frac{π}{4}, \frac{5 π}{4}$
and $- \frac{\cos θ}{\sqrt{1 + \cot^{2} θ}} = - \frac{\cos θ}{| \cos e c θ |} = - \sin θ \cos θ, w h e n θ \in (0, π)$
and $- 2 \tan θ \cot θ = - 2, w h e n θ \neq 0, \frac{π}{2}$
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, $θ \in (0, π) a n d θ \neq 0, \frac{π}{4}, \frac{5 π}{4}, \frac{π}{2}$
$∴ θ \in (0, π) - {\frac{π}{4}, \frac{π}{2}}$