If $\cos^{2} α - \sin^{2} α = \tan^{2} β,$ then $\tan^{2} α =$

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$\cos^{2} β + \sin^{2} β$

$\sin^{2} α - \sin^{2} β$

$\sin^{2} β - \cos^{2} β$

$\cos^{2} β - \sin^{2} β$

answer is D.

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Detailed Solution

We have,

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

On dividing both sides by $\cos^{2} α$ we get

$\begin{array}{l} 1 - \tan^{2} α = \frac{\tan^{2} β}{\cos^{2} α} = \tan^{2} β \sec^{2} α \\ 1 - \tan^{2} α = \tan^{2} β (1 + \tan^{2} α) \\ 1 - \tan^{2} α = \tan^{2} β + \tan^{2} α \tan^{2} β \\ 1 - \tan^{2} β = \tan^{2} α (1 + \tan^{2} β) \end{array}$

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

$∴ \tan^{2} α = \cos^{2} β - \sin^{2} β$

Alternate Method

We have,

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

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

Now, applying componendo and dividendo, we get

$\begin{matrix} \frac{\cos^{2} α - \sin^{2} α + \cos^{2} α + \sin^{2} α}{\cos^{2} α - \sin^{2} α - \cos^{2} α - \sin^{2} α} = \frac{\sin^{2} β + \cos^{2} β}{\sin^{2} β - \cos^{2} β} \\ \frac{2 \cos^{2} α}{- 2 \sin^{2} α} = \frac{1}{- (\cos^{2} β - \sin^{2} β)} \\ \frac{1}{\tan^{2} α} = \frac{1}{\cos^{2} β - \sin^{2} β} \end{matrix}$