If $a = \tan (α + β) - \tan (α - β), b = \tan (α + β) \tan (α - β)$ , then which of the following is/are Correct?

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$a = \frac{2 \sin 2 β}{\cos 2 α + \cos 2 β}$

$\tan (\frac{π}{4} + 2 β) = \frac{1 + a + b}{1 - a + b}$

$a = \frac{\sin 2 α}{\cos^{2} α - \sin^{2} β}$

$b = \frac{\sin^{2} α + \cos^{2} β - 1}{\cos^{2} α + \cos^{2} β - 1}$

answer is A, B, D.

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Detailed Solution

$a = \frac{\sin (α + β)}{\cos (α + β)} - \frac{\sin (α - β)}{\cos (α - β)} = \frac{2 \sin 2 β}{\cos 2 α + \cos 2 β}$

Also $a = \frac{\sin 2 β}{\cos^{2} α - \sin^{2} β}$

$\tan 2 β = \tan {(α + β) - (α - β)} = \frac{a}{1 + b} \frac{1 + \tan 2 β}{1 - \tan 2 β} = \frac{1 + a + b}{1 - a + b} \tan (\frac{π}{4} + 2 β) = \frac{1 + a + b}{1 - a + b} b = \frac{\sin (α + β)}{\cos (α + β)} . \frac{\sin (α - β)}{\cos (α - β)} = \frac{\sin^{2} α - \sin^{2} β}{\cos^{2} α - \cos^{2} β} = \frac{\sin^{2} α + \cos^{2} β - 1}{\cos^{2} α + \cos^{2} β - 1}$