If α−β is not odd multiple of π2 and sin⁡(α+β)cos⁡(α−β)=1−m1+m then prove that tan⁡π4−α=mtan⁡π4+β

Given sin⁡(α+β)cos⁡(α−β)=1−m1+mBy componendo & dividendo sin⁡(α+β)+cos⁡(α−β)sin⁡(α+β)−cos⁡(α−β)=1−m+1+m1−m−(1+m)⇒cos⁡π2−(α+β)+cos⁡(α−β)cos⁡π2−(α+β)−cos⁡(α−β)=21−m−1−m∵cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2⇒2cos⁡π2−α−β+α−β2cos⁡π2−α−β−α+β2−2sin⁡π2−α−β+α−β2sin⁡π2−α−β−α+β2=2−2m⇒cos⁡π4−βcos⁡π4−αsin⁡π4−βsin⁡π4−α=1m⇒cot⁡π4−βcot⁡π4−α=1m⇒tan⁡π4−βtan⁡π4−α=m⇒tan⁡π4−αcot⁡π2−π4−β=m⇒tan⁡π4−αcot⁡π4+β=m⇒tan⁡π4−α=mtan⁡π4+β. Hence proved.

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If $α - β$ is not odd multiple of $\frac{π}{2}$ and $\frac{\sin (α + β)}{\cos (α - β)} = \frac{1 - m}{1 + m}$ then prove that $\tan (\frac{π}{4} - α) = m \tan (\frac{π}{4} + β)$

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Detailed Solution

Given $\frac{\sin (α + β)}{\cos (α - β)} = \frac{1 - m}{1 + m}$

By componendo & dividendo

$\frac{\sin (α + β) + \cos (α - β)}{\sin (α + β) - \cos (α - β)} = \frac{1 - m + 1 + m}{1 - m - (1 + m)}$

$\Rightarrow \frac{\cos (\frac{π}{2} - (α + β)) + \cos (α - β)}{\cos (\frac{π}{2} - (α + β)) - \cos (α - β)} = \frac{2}{1 - m - 1 - m}$

$(∵ \cos C + \cos D = 2 \cos (\frac{C + D}{2}) \cos (\frac{C - D}{2}))$

$\Rightarrow \frac{2 \cos (\frac{\frac{π}{2} - α - β + α - β}{2}) \cos (\frac{\frac{π}{2} - α - β - α + β}{2})}{- 2 \sin (\frac{\frac{π}{2} - α - β + α - β}{2}) \sin (\frac{\frac{π}{2} - α - β - α + β}{2})}$ $= \frac{2}{- 2 m}$