If sin⁡(θ+α)cos⁡(θ−α)=1−m1+m, then tan⁡π4−θ=

1−m1+m=cos⁡90∘−θ−αcos⁡(θ−α) ⇒1+m+1−m1+m−1+m=cos⁡(θ−α)+cos⁡90∘−θ−αcos⁡(θ−α)−cos⁡90∘−θ−α ⇒22m=2cos⁡45∘−α⋅cos⁡45∘−θ2sin⁡45∘−α⋅sin⁡45∘−θ⇒1m=cot⁡45∘−αcot⁡45∘−θ⇒tan⁡(π/4−θ)=mcot⁡(π/4−α)

If sin⁡(θ+α)cos⁡(θ−α)=1−m1+m, then tan⁡π4−θ=

1−m1+m=cos⁡90∘−θ−αcos⁡(θ−α) ⇒1+m+1−m1+m−1+m=cos⁡(θ−α)+cos⁡90∘−θ−αcos⁡(θ−α)−cos⁡90∘−θ−α ⇒22m=2cos⁡45∘−α⋅cos⁡45∘−θ2sin⁡45∘−α⋅sin⁡45∘−θ⇒1m=cot⁡45∘−αcot⁡45∘−θ⇒tan⁡(π/4−θ)=mcot⁡(π/4−α)

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$If \frac{\sin (θ + α)}{\cos (θ - α)} = \frac{1 - m}{1 + m}, then \tan (\frac{π}{4} - θ) =$

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$mtan (\frac{π}{4} + α)$

$msin (\frac{π}{4} - α)$

$mcot (\frac{π}{4} - α)$

$mcos (\frac{π}{4} + α)$

answer is D.

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Detailed Solution

$\frac{1 - m}{1 + m} = \frac{\cos (90^{\circ} - θ - α)}{\cos (θ - α)} \Rightarrow \frac{1 + m + 1 - m}{1 + m - 1 + m} = \frac{\cos (θ - α) + \cos (90^{\circ} - θ - α)}{\cos (θ - α) - \cos (90^{\circ} - θ - α)} \begin{array}{l} \Rightarrow \frac{2}{2 m} = \frac{2 \cos (45^{\circ} - α) \cdot \cos (45^{\circ} - θ)}{2 \sin (45^{\circ} - α) \cdot \sin (45^{\circ} - θ)} \\ \Rightarrow \frac{1}{m} = \cot (45^{\circ} - α) \cot (45^{\circ} - θ) \\ \Rightarrow \tan (π / 4 - θ) = mcot (π / 4 - α) \end{array}$