Tangent subtraction formula

The formula for subtracting tangents is as follows: $$ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} $$

This formula is valid as long as the difference a-b does not equal ±90° + k·180°, where k is an integer.

$$ \alpha - \beta \ne \frac{\pi}{2}+ k \cdot \pi \ \ \ \ \ k \in Z $$

This is because the tangent is undefined at these points.

A practical example

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

$$ \tan a = \tan 60° = \sqrt{3} $$

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

The tangent of the difference a-b is the tangent of 30°.

$$ \tan (a-b) = \tan(60°-30°) = \tan(30°) = \frac{\sqrt{3}}{3} $$

Note: The tangent of the difference between two angles a-b is not simply the difference between their tangents. $$ \tan(60°-30°) = \tan(30°) = \frac{\sqrt{3}}{3} \ne \tan(60°) - \tan(30°) = \sqrt{3} - \frac{\sqrt{3}}{3} = \frac{3\sqrt{3} - \sqrt{3}}{3} = \frac{2\sqrt{3}}{3} $$

To subtract these two angles, we use the tangent subtraction formula.

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

Where a = 60° and b = 30°.

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

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

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

$$ \tan (60°-30°) = \frac{ \frac{2\sqrt{3}}{3} }{2} $$

$$ \tan (60°-30°) = \frac{2\sqrt{3}}{3} \cdot \frac{1}{2} $$

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

The result is correct:

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

The proof

To prove the formula:

$$ \tan(\alpha - \beta) $$

we rewrite the subtraction as an addition:

$$ \tan(\alpha - \beta) = \tan[\alpha + (- \beta)] $$

Next, we apply the tangent addition formula:

$$ \tan(\alpha - \beta) = \tan[\alpha + (- \beta)] = \frac{ \tan \alpha + \tan(- \beta) }{ 1 - \tan \alpha \tan(- \beta) } $$

Since the tangent is an odd function, tan(-β) = -tan(β).

$$ \tan(\alpha - \beta) = \frac{ \tan \alpha - \tan \beta }{ 1 - \tan \alpha [ - \tan(\beta) ] } $$

$$ \tan(\alpha - \beta) = \frac{ \tan \alpha - \tan \beta }{ 1 + \tan \alpha \tan \beta } $$

This completes the proof of the initial formula.

And that’s how it’s done.