Tangent Addition Formula

The formula for the tangent of the sum of two angles is: $$ \tan(\alpha + \beta) = \frac{ \tan \alpha + \tan \beta }{1- \tan \alpha \tan \beta } $$

It's important to remember that the tangent of the sum of two angles is not simply the sum of the tangents of the individual angles:

$$ \tan(\alpha + \beta) \ne \tan \alpha + \tan \beta $$

Note: This formula is valid within the domain of the tangent function. Therefore, the sum of the angles α+β must not be equal to ±90° + k·180°, where k is any integer. In radians, the condition is: $$ \alpha + \beta \ne \frac{\pi}{2} + k \cdot \pi \ \ \ \ \forall \ k \in Z $$

A Practical Example

Let’s take the tangents of two angles, a=30° and b=60°:

$$ \tan a = \tan 30° = \frac{\sqrt{3}}{3} $$

$$ \tan b = \tan 60° = \sqrt{3} $$

The tangent of their sum, a+b, is the tangent of 90°, which is undefined.

$$ \tan (a+b) = \tan(30°+60°) = \tan(90°) = \nexists $$

Note: However, if we add the individual tangents, we get a finite value: $$ \tan a + \tan b = \tan 30° + \tan 60° = \frac{\sqrt{3}}{3} + \sqrt{3} = \frac{\sqrt{3}+3 \sqrt{3}}{3} $$ This demonstrates what was mentioned earlier: the sum of the tangents of two angles is not equal to the tangent of the sum of the angles. $$ a + b \ne \frac{\pi}{2} + k \cdot \pi \ \ \ \ \forall \ k \in Z $$

To find the tangent of the sum of these angles, we need to use the tangent addition formula:

$$ \tan (a+b) = \frac{ \tan a + \tan b }{1-\tan a \tan b} $$

Where a=30° and b=60°.

$$ \tan (30°+60°) = \frac{ \tan 30° + \tan 60° }{1-\tan 30° \tan 60°} $$

$$ \tan (30°+60°) = \frac{ \frac{\sqrt{3}}{3} + \sqrt{3} }{1-\frac{\sqrt{3}}{3} \cdot \sqrt{3}} $$

$$ \tan (30°+60°) = \frac{ \frac{\sqrt{3} + 3\sqrt{3}}{3}}{1-\frac{3}{3}} $$

$$ \tan (30°+60°) = \frac{ \frac{\sqrt{3} + 3\sqrt{3}}{3}}{1-1} $$

$$ \tan (30°+60°) = \frac{ \frac{\sqrt{3} + 3\sqrt{3}}{3}}{0} $$

This is division by zero, which means the tangent is undefined.

$$ \tan (30°+60°) = \nexists $$

Therefore, the result is correct.

The Proof

The tangent of an angle x is the ratio of the sine to the cosine of that angle:

$$ \tan x = \frac{\sin x}{\cos x} $$

Now, let's consider the angle x = a+b, the sum of two angles a and b:

$$ \tan (a+b) = \frac{\sin (a+b)}{\cos (a+b)} $$

We apply the sine addition formula to the numerator:

$$ \tan (a+b) = \frac{\sin a \cos b + \sin b \cos a}{\cos (a+b)} $$

And we apply the cosine addition formula to the denominator:

$$ \tan (a+b) = \frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b - \sin a \sin b} $$

Now, divide both the numerator and denominator by cos(a)·cos(b):

$$ \tan (a+b) = \frac{ \frac{ \sin a \cos b + \sin b \cos a }{ \cos a \cos b } }{ \frac{ \cos a \cos b - \sin a \sin b }{ \cos a \cos b } } $$

$$ \tan (a+b) = \frac{ \frac{ \sin a \cos b}{ \cos a \cos b } + \frac{\sin b \cos a }{ \cos a \cos b }}{ \frac{ \cos a \cos b }{ \cos a \cos b } - \frac{ \sin a \sin b }{ \cos a \cos b } } $$

$$ \tan (a+b) = \frac{ \frac{ \sin a}{ \cos a } + \frac{\sin b }{ \cos b } }{1 - \frac{ \sin a }{ \cos a } \cdot \frac{ \sin b }{\cos b } } $$

Since we know that tan(a) = sin(a)/cos(a):

$$ \tan (a+b) = \frac{ \tan a + \frac{\sin b }{ \cos b } }{1 - \tan a \cdot \frac{ \sin b }{\cos b } } $$

And that tan(b) = sin(b)/cos(b):

$$ \tan (a+b) = \frac{ \tan a + \tan b }{1 - \tan a \cdot \tan b } $$

This gives us the formula we wanted to prove.

And so on.