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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

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    $\begingroup$ I would say the Pythagorean theorem/parametrization of the unit circle is the deep reason, and almost every basic trigonometric identity boils down to this. $\endgroup$
    – peek-a-boo
    May 14, 2021 at 2:12
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    $\begingroup$ You don't even need to use trigonometry here, really; you can write $x^2 + y^2 = 1 \implies 1 - y^2 = x^2 \implies 1/x^2 - y^2 = 1 \implies (1/x- y)(1/x + y) = 1$. $\endgroup$ May 14, 2021 at 2:13

2 Answers 2

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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.

enter image description here

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, $O$.

enter image description here

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.)

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  • $\begingroup$ perfect. thank you $\endgroup$ May 14, 2021 at 2:59
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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 $$ \dfrac {1}{x} = {\sec \theta-\tan \theta}.$$

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