If 1cos⁡αcos⁡β+tan⁡αtan⁡β=tan⁡γ, 0<α,β<π then 1−tan2⁡γ<0 for

We have 1−tan2⁡γ=cos2⁡αcos2⁡β−(1+sin⁡αsin⁡β)2cos2⁡αcos2⁡β=1−sin2⁡α1−sin2⁡β−1+2sin⁡αsin⁡β+sin2⁡αsin2⁡βcos2⁡αcos2⁡β=−(sin⁡α+sin⁡β)2cos2⁡αcos2⁡β<0(sin⁡α+sin⁡β≠0, as 0<α,β<π)

If 1cos⁡αcos⁡β+tan⁡αtan⁡β=tan⁡γ, 0<α,β<π then 1−tan2⁡γ<0 for

We have 1−tan2⁡γ=cos2⁡αcos2⁡β−(1+sin⁡αsin⁡β)2cos2⁡αcos2⁡β=1−sin2⁡α1−sin2⁡β−1+2sin⁡αsin⁡β+sin2⁡αsin2⁡βcos2⁡αcos2⁡β=−(sin⁡α+sin⁡β)2cos2⁡αcos2⁡β<0(sin⁡α+sin⁡β≠0, as 0<α,β<π)

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If $\frac{1}{\cos α \cos β} + \tan α \tan β = \tan γ$ , $0 < α, β < π$ then $1 - \tan^{2} γ < 0$ for

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all values of $α$ and $β$

no values of $α$ and $β$

finite number of values of $α$ and $β$

infinite number of values of $α$ and $β$

answer is A.

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Detailed Solution

We have $1 - \tan^{2} γ =$

$\frac{\cos^{2} α \cos^{2} β - (1 + \sin α \sin β)^{2}}{\cos^{2} α \cos^{2} β}$

$\begin{array}{l} = \frac{(1 - \sin^{2} α) (1 - \sin^{2} β) - (1 + 2 \sin α \sin β + \sin^{2} α \sin^{2} β)}{\cos^{2} α \cos^{2} β} \\ = \frac{- (\sin α + \sin β)^{2}}{\cos^{2} α \cos^{2} β} < 0 (\sin α + \sin β \neq 0, as 0 < α, β < π) \end{array}$