Sums and differences of arctangents

What you will be able to do

• add two inverse tangents with $\dfrac{x + y}{1 - xy}$ and decide when a correction of $\pm\pi$ is needed;
• subtract inverse tangents with $\dfrac{x - y}{1 + xy}$;
• prove the rule and use it to combine three inverse tangents.

Starting from the tangent of a sum

Let $A = \tan^{-1}x$ and $B = \tan^{-1}y$. Both angles lie in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. The addition formula for the tangent gives

It is tempting to say that $A + B = \tan^{-1}\dfrac{x + y}{1 - xy}$. But the inverse tangent returns an angle in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, and $A + B$ can be anywhere in $(-\pi, \pi)$. The equation is true only when $A + B$ is in the branch. The product $xy$ tells us when.

The rule

There are three cases.

• If $xy < 1$: $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x + y}{1 - xy}$.
• If $xy > 1$ and $x > 0$: $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\dfrac{x + y}{1 - xy}$.
• If $xy > 1$ and $x < 0$: $\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\dfrac{x + y}{1 - xy}$.

If $xy = 1$ the fraction is not defined. Then the sum is $\dfrac{\pi}{2}$ when $x > 0$, and $-\dfrac{\pi}{2}$ when $x < 0$.

Why. By the cosine addition formula, $\cos(A + B) = \cos A\cos B\,(1 - xy)$. Here $\cos A$ and $\cos B$ are positive. So if $xy < 1$ then $\cos(A + B) > 0$, which puts $A + B$ in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, the branch. If $xy > 1$ then $\cos(A + B) < 0$, and $A + B$ is beyond the branch. The angle $\pi$ brings it back, because the tangent has period $\pi$.

The difference. Replace $y$ by $-y$, using $\tan^{-1}(-y) = -\tan^{-1}y$:

A worked example

The picture shows part (a) below: two angles from the same corner add up to half of a right angle, which is $\dfrac{\pi}{4}$.

Find (a) $\tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3}$, (b) $\tan^{-1}2 + \tan^{-1}3$, (c) $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$, (d) $\tan^{-1}3 - \tan^{-1}2$.

1. (a) $xy = \dfrac{1}{6} < 1$. The fraction is $\dfrac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \dfrac{5/6}{5/6} = 1$. The sum is $\tan^{-1}1 = \dfrac{\pi}{4}$.
2. (b) $xy = 6 > 1$ and $x > 0$. The fraction is $\dfrac{5}{1 - 6} = -1$. The sum is $\pi + \tan^{-1}(-1) = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$.
3. (c) Use (b), then add $\tan^{-1}1 = \dfrac{\pi}{4}$. The total is $\dfrac{3\pi}{4} + \dfrac{\pi}{4} = \pi$.
4. (d) $xy = 6 > -1$, so the difference is $\tan^{-1}\dfrac{3 - 2}{1 + 6} = \tan^{-1}\dfrac{1}{7}$.

Practise with new numbers

Try it in Python

The program adds pairs of inverse tangents directly and with the formula. The last column shows how many times $\pi$ the formula is away from the truth.

Common mistakes

• Using the formula without looking at $xy$. If $xy > 1$ the answer is off by $\pi$.
• A plus sign in the denominator of a sum. For the sum it is $1 - xy$. For the difference it is $1 + xy$.
• Adding $\pi$ for two negative numbers. For $x < 0$ and $xy > 1$ subtract $\pi$.
• Forgetting the case $xy = 1$. The fraction has a zero denominator, and the sum is $\pm\dfrac{\pi}{2}$.

Remember these

• $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x + y}{1 - xy}$ when $xy < 1$.
• When $xy > 1$ add $\pi$ if $x > 0$, and subtract $\pi$ if $x < 0$. When $xy = 1$ the sum is $\pm\dfrac{\pi}{2}$.
• $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\dfrac{x - y}{1 + xy}$ when $xy > -1$.
• For three terms, combine two, check the branch, then add the third.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\tan^{-1}\dfrac{1}{4} + \tan^{-1}\dfrac{3}{5}$.

2. Find $\tan^{-1}4 + \tan^{-1}\dfrac{1}{4}$.

3. Find $\tan^{-1}(-2) + \tan^{-1}(-3)$.

4. Find $\tan^{-1}2 - \tan^{-1}\dfrac{1}{3}$.

5. A student writes $\tan^{-1}5 + \tan^{-1}6 = \tan^{-1}\left(-\dfrac{11}{29}\right)$. Why can this not be right?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.