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

• 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. May 14 '21 at 2:12
• You don't even need to use trigonometry here, really; you can write $x^2 + y^2 = 1 \implies 1 - y^2 = x^2 \implies 1/x^2 - y^2 = 1 \implies (1/x- y)(1/x + y) = 1$. May 14 '21 at 2:13

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.)

• perfect. thank you May 14 '21 at 2:59

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 $$\dfrac {1}{x} = {\sec \theta-\tan \theta}.$$