A simplified version of simple question

I don't see how the identity below is related to $(\tan \theta + \sec \theta)(\tan \alpha + \sec \alpha) = 1$ .

Using the identity $\sec^2 x-\tan^2 x=1$ , we can conclude that $\sec \theta =\sec x, \tan \theta =-\tan x$

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$\begin{aligned} (\tan \theta + \sec \theta) (\tan \alpha + \sec \alpha) & = 1 \\ \left(\frac {\sin \theta + 1}{\cos \theta} \right) \left(\frac {\sin \alpha + 1}{\cos \alpha} \right) & = 1 \\ \left(\frac {\sin \theta + 1}{\cos \theta} \right) & = \left(\frac{\cos \alpha} {\sin \alpha + 1} \right) & \small \blue{\text{Using half-angle tangent substitution}} \\ \frac {\frac {2\tan \frac \theta 2}{1+\tan^2 \frac \theta 2}+1}{\frac {1-\tan^2 \frac \theta 2}{1+\tan^2 \frac \theta 2}} & = \frac {\frac {1-\tan^2 \frac \alpha 2}{1+\tan^2 \frac \alpha 2}}{\frac {2\tan \frac \alpha 2}{1+\tan^2 \frac \alpha 2}+1} \\ \frac {(1+\tan \frac \theta 2)^2}{1-\tan^2 \frac \theta 2} & = \frac {1-\tan^2 \frac \alpha 2} {(1+\tan \frac \alpha 2)^2} \\ \frac {1+\tan \frac \theta 2}{1-\tan \frac \theta 2} & = \frac {1-\tan \frac \alpha 2} {1+\tan \frac \alpha 2} \\ \tan \left(\frac \pi 4 + \frac \theta 2 \right) & = \tan \left(\frac \pi 4 - \frac \alpha 2 \right) \\ \implies \alpha & = - \theta \end{aligned}$

Therefore,

$\begin{aligned} \tan^2 \theta - \sec^2 \alpha & = \tan^2 \theta - \sec^2 (-\theta) \\ & = \tan^2 \theta - \sec^2 \theta \\ & = \tan^2 \theta - (1 + \tan^2 \theta ) \\ & = \boxed{-1} \end{aligned}$

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Reference: Half-angle tangent substitution

Using the identity $\sec^2 x-\tan^2 x=1$ , we can conclude that $\sec \theta =\sec x, \tan \theta =-\tan x$

Hence, $\tan^2 \theta -\sec^2 x=-(\sec^2 x-\tan^2 x) =\boxed {-1}$ .