If $α, β$ and $γ$ are connected by the relation
$2 \tan^{2} α \tan^{2} β \tan^{2} γ + \tan^{2} α \tan^{2} β + \tan^{2} β \tan^{2} γ + \tan^{2} γ \tan^{2} α = 1 t h e n$

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$\sin^{2} α + \sin^{2} β + \sin^{2} γ = 1$

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 2$

$\cos 2 α + \cos 2 β + \cos 2 γ = 1$

$\cos (α + β) \cos (α - β) = \cos^{2} γ$

answer is A, B, C.

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Detailed Solution

$\sum \tan^{2} β \tan^{2} γ = 1 - 2 \tan^{2} α \tan^{2} β \tan^{2} γ \sin^{2} α + \sin^{2} β + \sin^{2} γ = \frac{\tan^{2} α}{1 + \tan^{2} α} + \frac{\tan^{2} β}{1 + \tan^{2} β} + \frac{\tan^{2} γ}{1 + \tan^{2} γ} \frac{\sum \tan^{2} α (1 + \tan^{2} β + \tan^{2} γ + \tan^{2} β \tan^{2} γ)}{1 + \sum \tan^{2} α + \sum \tan^{2} β \tan^{2} γ + \tan^{2} α \tan^{2} β \tan^{2} γ} \frac{\sum \tan^{2} α + 2 \sum \tan^{2} α \tan^{2} β + 3 \tan^{2} α \tan^{2} β \tan^{2} γ}{1 + \sum \tan^{2} α + 1 - \tan^{2} α \tan^{2} β \tan^{2} γ} = \frac{2 + \sum \tan^{2} α - \tan^{2} α \tan^{2} β \tan^{2} γ}{2 + \sum \tan^{2} α - \tan^{2} α \tan^{2} β \tan^{2} γ} = 1 \begin{array}{l} \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 2 \\ \Rightarrow \cos 2 α + \cos 2 β + \cos 2 γ = 1 \\ \Rightarrow 2 \cos (α + β) \cos (α - β) = 2 \sin^{2} γ \end{array}$
or

$\tan^{2} α = \tan^{2} β = \tan^{2} γ = \frac{1}{2} s a t i s f y t h e g i v e n r e l a t i o n S o \sin^{2} α + \sin^{2} β + \sin^{2} γ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$