Let S={θ∈[0,2π):tan⁡(πcos⁡θ)+tan⁡(πsin⁡θ)=0} then ∑θ∈S sin2⁡θ+π4 is equal to ________

tan⁡(πcos⁡θ)=−tan⁡(πsin⁡θ) =tan⁡(π−πsin⁡θ)⇒πcos⁡θ+πsin⁡θ=π⇒cos⁡θ+sin⁡θ=1⇒sin⁡θ+π4=12∴θ+π4=nπ+(−1)nπ4,n∈Zθn=nπ+(−1)nπ4−π4,n∈Zn=0⇒θ=0n=1⇒θ1=π−π4−π4=π2n=1⇒θ1=2π+π4−π4=2π∉[0,2π]∴θ=0,π2∈[0,2π)∑θ∈S sin2⁡θ+π4=sin2⁡0+π4+sin2⁡π2+π4=sin2⁡π4+cos2⁡π4=1

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Let $S = {θ \in [0, 2 π) : \tan (πcos θ) + \tan (πsin θ) = 0}$ then $\sum_{θ \in S} \sin^{2} (θ + \frac{π}{4})$ is equal to ________

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Detailed Solution

$\begin{array}{l} \tan (πcos θ) = - \tan (πsin θ) \\ = \tan (π - πsin θ) \\ \Rightarrow πcos θ + πsin θ = π \\ \Rightarrow \cos θ + \sin θ = 1 \\ \Rightarrow \sin (θ + \frac{π}{4}) = \frac{1}{\sqrt{2}} \\ ∴ θ + \frac{π}{4} = nπ + (- 1)^{n} \frac{π}{4}, n \in Z \\ θ_{n} = nπ + (- 1)^{n} \frac{π}{4} - \frac{π}{4}, n \in Z \end{array}$
$\begin{array}{l} n = 0 \Rightarrow θ = 0 \\ n = 1 \Rightarrow θ_{1} = π - \frac{π}{4} - \frac{π}{4} = \frac{π}{2} \\ n = 1 \Rightarrow θ_{1} = 2 π + \frac{π}{4} - \frac{π}{4} = 2 π \notin [0, 2 π] \\ ∴ θ = 0, \frac{π}{2} \in [0, 2 π) \\ \sum_{θ \in S} \sin^{2} (θ + \frac{π}{4}) = \sin^{2} (0 + \frac{π}{4}) + \sin^{2} (\frac{π}{2} + \frac{π}{4}) \\ = \sin^{2} \frac{π}{4} + \cos^{2} \frac{π}{4} = 1 \end{array}$