If cot⁡(θ−α), 3cot⁡θ, cot⁡(θ+α)are in AP (where, θ≠nπ2, α≠kπ, k∈I) then 2sin2⁡θsin2⁡αis equal to ______

we have 6cot⁡θ=cot⁡(θ−α)+cot⁡(θ+α)⇒6cos⁡θsin⁡θ=sin⁡2θsin⁡(θ−α)+sin⁡(θ+α)⇒6cos⁡θsin⁡θ=2sin⁡θcos⁡θsin2⁡θ−sin2⁡α⇒3sin2⁡θ−sin2⁡α=sin2⁡θ or 2sin2⁡θ=3sin2⁡α⇒2sin2⁡θsin2⁡α=3

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If $\cot (θ - α), 3 \cot θ, \cot (θ + α)$ are in AP (where, $θ \neq \frac{nπ}{2}, α \neq kπ, k \in I$ ) then $\frac{2 \sin^{2} θ}{\sin^{2} α}$ is equal to ______

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answer is 3.

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Detailed Solution

we have $6 \cot θ = \cot (θ - α) + \cot (θ + α)$
$\begin{array}{l} \Rightarrow \frac{6 \cos θ}{\sin θ} = \frac{\sin 2 θ}{\sin (θ - α) + \sin (θ + α)} \\ \Rightarrow \frac{6 \cos θ}{\sin θ} = \frac{2 \sin θcos θ}{\sin^{2} θ - \sin^{2} α} \\ \Rightarrow 3 (\sin^{2} θ - \sin^{2} α) = \sin^{2} θ \\ or 2 \sin^{2} θ = 3 \sin^{2} α \Rightarrow \frac{2 \sin^{2} θ}{\sin^{2} α} = 3 \end{array}$