Prove that $\alpha + \beta = \gamma$, given the values of $\tan \alpha$, $\tan \beta$ and $\tan \gamma$

Question

This is a question from my Mathematics textbook.

The given values of $\tan \alpha$, $\tan \beta$ and $\tan \gamma$ are :

$\tan \alpha = \dfrac{1}{\sqrt{x(x^2+x+1)}}$

$\tan \beta = \dfrac{\sqrt{x}}{\sqrt{x^2+x+1}}$

$\tan \gamma = \sqrt{x^{-3}+x^{-2}+x^{-1}}$

The following is a compound angle identity in trigonometry : $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$ We can substitute the values of $\tan \alpha$ and $\tan \beta$ in the above mentioned identity and obtain the value of $\tan (\alpha + \beta)$, which, on simplifying, will give us $\sqrt{x^{-3}+x^{-2}+x^{-1}}$ which is the value of $\tan \gamma$. So, basically, we deduce that $\tan (\alpha + \beta) = \tan \gamma$

My main question here is that if we have $\tan (\alpha + \beta) = \tan \gamma$, it does not necessarily imply that $\alpha + \beta = \gamma$, since $\forall n \in \Bbb Z, \tan (n\pi + \theta) = \tan \theta$, which can be understood by taking into account the fact that $\tan$ is a periodic function with $P = \pi$.

So, in my opinion, some condition should have been mentioned in the question, for example, $\alpha + \beta < \pi$ and $\gamma < \pi$, so that $\tan (\alpha + \beta) = \tan \gamma$ would imply that $\alpha + \beta = \gamma$.

Let me know if I'm right, thanks!

Answer 1

The domain and range are between $(\infty, -\infty)$ and $ \tan(x)$ is a monotonic function.

So with $$ \tan \alpha =A, \;\tan \beta =B,\;\tan \gamma =C\;$$ so long as the relation

$$ C= \dfrac{A+B}{1-AB},$$

is satisfied, it is an unconditional identity.

This is satisfied and the relation always holds true across all asymptotes in the domains.