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
Best Answer
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.)