If tan⁡(πsin⁡θ)=cot⁡(πcos⁡θ), then sin⁡θ+π4=

tan⁡(πsin⁡θ)=tan⁡π2−πcos⁡θ πsin⁡θ=nπ+π2−πcos⁡θ sinθ+cosθ=2n+12 n=0,-1 12sinθ+12cosθ=±122 sinθ+π4=±122

If tan⁡(πsin⁡θ)=cot⁡(πcos⁡θ), then sin⁡θ+π4=

tan⁡(πsin⁡θ)=tan⁡π2−πcos⁡θ πsin⁡θ=nπ+π2−πcos⁡θ sinθ+cosθ=2n+12 n=0,-1 12sinθ+12cosθ=±122 sinθ+π4=±122

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If $\tan (πsin θ) = \cot (πcos θ), then \sin (θ + \frac{π}{4}) =$

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$\frac{1}{\sqrt{2}}$

$\pm \frac{1}{2 \sqrt{2}}$

$\pm \frac{1}{\sqrt{2}}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$\tan (πsin θ) = \tan (\frac{π}{2} - πcos θ) πsin θ = nπ + \frac{π}{2} - πcos θ sinθ + cosθ = \frac{2 n + 1}{2} n = 0, - 1 \frac{1}{\sqrt{2}} sinθ + \frac{1}{\sqrt{2}} cosθ = \frac{\pm 1}{2 \sqrt{2}} \sin (θ + \frac{π}{4}) = \pm \frac{1}{2 \sqrt{2}}$