Mark44 said:
The given equations are ##\tan(\alpha) = \frac a {a + 1}## and ##\tan(\beta) = \frac 1 {2a + 1}##. These lead to the equation ##\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}##.
As you've seen, ##\alpha + \beta = \frac \pi 4## is a solution, but there are an infinite number of solutions to this equation above, one of which is ##-\frac{3\pi} 4##, which is smaller than ##\frac \pi 4##. There are many others that are even more negative; i.e., smaller. This is why I said that the problem statement isn't clear, and would have been improved by asking for the smallest positive value of ##\alpha + \beta##.
I beg to AGREE with Mark on this point.
In general, when using Inverse trig functions to solve trig equations, we must exercise care.
The solution to an equation such as ##\displaystyle \ \tan(\theta) = u\ ## is ##\displaystyle \ \theta = \arctan(u) +k\pi \, ,\ ## where ##k## is an integer.
Of course, once you introduce the inverse tangent, whose principal domain is ##(\frac{-\pi}2, \frac \pi 2)##, then there is only one solution.
As a general method, the following is not particularly good at getting an overall solution in closed form, but it does use the principal domain of ##\displaystyle \arctan \, .##
Alternate approach:
If ##\displaystyle \tan(\alpha) = \dfrac a {a + 1} \, , \ ## then using ##\displaystyle \arctan \ ##
we have ##\displaystyle \alpha = \arctan(\tan(\alpha)) = \arctan \, \dfrac a {a + 1} \, . \ ##
Similarly, ##\displaystyle \beta = \arctan(\tan(\beta)) = \arctan \, \dfrac {1} {2a + 1} \, . \ ##
For ##\displaystyle -1 < a < \frac {-1} 2 \, ,## each of the above expressions gives a negative result for ##\alpha## and for ##\beta##.
Using any particular value for ##a## in this interval gives a value of ##\displaystyle \, \frac {-3\pi} 4 \, ## for ##\displaystyle \alpha + \beta =\arctan \, \dfrac a {a + 1} + \arctan \, \dfrac {1} {2a + 1} \, . ##
For example, try ##\displaystyle a=\frac{-3}{4} \ ## on a scientific calculator. .
