Why are $\frac{1}{\cos \theta}+\tan{\theta}$ and $\frac{1}{\cos\theta}-\tan{\theta}$ always reciprocals (besides simply multiplying to get $1$)? Can anybody help me understand why these terms are always reciprocals? (theta <= 45°)
$$ x =  \frac{1}{\cos \theta} + \tan{\theta} $$
$$ \frac{1}{x} =  \frac{1}{\cos \theta} - \tan{\theta} $$
I understand that if we multiply them, they equal $1$ (because of the equation for a circle).
$$\begin{align}
1 &= (\frac{1}{\cos \theta} + \tan{θ})(\frac{1}{\cos \theta} - \tan{θ}) \\[4pt]
1 &= \frac{1}{(\cos{\theta})^{2}} - \frac{\tan{\theta}}{\cos{\theta}} +  \frac{\tan{\theta}}{\cos{\theta}} - (\tan{\theta})^2 \\[4pt]
1 &= \frac{1}{(\cos{\theta})^2} - (\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\cos{\theta})^2(\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\sin{\theta})^2 
\end{align}$$
But I am looking for a deeper understanding? Regards

Proof:
If $$x = \sec \theta + \tan \theta \ (\text {here} \sec \theta = \dfrac {1}{\cos \theta})$$
Then $$ \dfrac {1}{x} = \dfrac {1}{\sec \theta+\tan \theta}$$
Rationalizing the denominator by multiplying by $\sec \theta - \tan \theta$ we have $$ \dfrac {1}{x} = \dfrac {\sec \theta-\tan \theta}{\sec^2 \theta - \tan^2 \theta}$$
As $\sec^2 \theta - \tan^2 \theta = 1$ via the Pythagorean theorem, we then have $$ \dfrac {1}{x} = {\sec \theta-\tan \theta}.$$
