Interesting Trigonometry Identity Prelude
While studying trigonometry, I came across this very interesting problem. It wasn't very difficult to solve, however it's result was quite interesting. I have given the solution below. Try to solve it yourself before looking.

Problem
The diagram below shows three equal squares, with angels $\alpha$, $\beta$, $\gamma$ as marked. Prove that $ \alpha + \beta = \gamma $.


Hint
If you are not in the midst of a trigonometry course, chances are you will need to see this hint, as it is an uncommon identity.
Hover below to see a hint.

 $ \arctan a + \arctan b = \arctan{\frac{a + b}{1 - ab}} $

This identity was solved earlier in the book I found this problem in (see Postscript). Try to prove this identity for yourself, too!

Postscript
This problem came from the textbook: "Trigonometry" by I. M. Gelfand.


Proof

Step 1: Gather information from diagram.


*

*$ \alpha = \arctan \frac{1}{3} $

*$ \beta = \arctan \frac{1}{2} $

*$ \gamma = \arctan 1 $



Step 2: Add $\alpha$ and $\beta$ and simplify.
$ \alpha + \beta = \arctan \frac{1}{3} + \arctan \frac{1}{2} = $
...Now, using an identity (see hint given in the question)...
$ = \arctan {\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3}\frac{1}{2}}}
= \arctan {\frac{\frac{5}{6}}{1 - \frac{1}{6}}}
= \arctan {\frac{\frac{5}{6}}{\frac{5}{6}}}
= \arctan 1
= \gamma $

Answer
Indeed,  $$ \alpha + \beta = \gamma $$
