If tanπcosθ=cotπsinθ then cosθ-π4=

tan⁡[πcos⁡θ]=cot⁡[πsin⁡θ]tan⁡[πcos⁡θ]=tan⁡π2−πsin⁡θπcos⁡θ=nπ+π2−πsin⁡θcos⁡θ+sin⁡θ=±12 (∵n=0,-1)12cos⁡θ+12sin⁡θ=±122cos⁡θ−π4=±122

If tanπcosθ=cotπsinθ then cosθ-π4=

tan⁡[πcos⁡θ]=cot⁡[πsin⁡θ]tan⁡[πcos⁡θ]=tan⁡π2−πsin⁡θπcos⁡θ=nπ+π2−πsin⁡θcos⁡θ+sin⁡θ=±12 (∵n=0,-1)12cos⁡θ+12sin⁡θ=±122cos⁡θ−π4=±122

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$If \tan (πcosθ) = \cot (πsinθ) then \cos (θ - \frac{π}{4}) =$

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

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

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

$\pm \frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} \tan [π \cos θ] = \cot [π \sin θ] \\ \tan [π \cos θ] = \tan [\frac{π}{2} - π \sin θ] \\ π \cos θ = n π + \frac{π}{2} - π \sin θ \\ \cos θ + \sin θ = \pm \frac{1}{2} (∵ n = 0, - 1) \\ \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ = \pm \frac{1}{2 \sqrt{2}} \\ \cos [θ - \frac{π}{4}] = \pm \frac{1}{2 \sqrt{2}} \end{array}$