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
 A: There is a theorem (or set of theorems) of geometry called the
Power of a Point.
Note that this theorem is easily proved without using any trigonometry.

A particular case of the power of a point says that if you have a line through the point $A$ that is tangent to a circle at $B,$
and another line through $A$ that intersects the same circle at $C$ and $D,$
as shown in the figure above, then
$$ AB^2 = AC \cdot AD. $$
Now lets add some more specific properties to the figure. First, suppose we take the distance $AB$ as our unit of length, so $AB = 1.$
Next, suppose $\angle BAC = \theta.$
Finally, suppose the line $AD$ passes through the center of the circle, $O$.

Now $BO = \tan \theta$ and $AO = \frac{1}{\cos\theta} = \sec \theta$.
Observe that $AC = \sec\theta - \tan\theta$ and that $AD = \sec\theta + \tan\theta$.
Recalling the formula for the power of a point in a case like this,
$AB^2 = AC \cdot AD,$
and putting the particular lengths of the segments in this example into that formula,
we find that
$$ 1^2 = (\sec\theta - \tan\theta)(\sec\theta + \tan\theta), $$
and therefore
$$ \sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta}. $$
So your trigonometric identity is simply a special case of the power of a point.
Note that it is not restricted only to $0 \leq \theta \leq 45^\circ.$
The geometric theorem shows that the identity is true for any acute angle.
If you use the usual definition of trigonometric functions for angles outside the range from zero to a right angle, the identity is good for all angles for which the cosine is not zero. But that takes a bit more interpretation if you want to make something geometric out of it. (In particular, you have to deal with the fact that the cosine and tangent are sometimes negative.)
A: 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 $$ \dfrac {1}{x} = {\sec \theta-\tan \theta}.$$
