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

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!

• I think you should be correct—if $\alpha$, $\beta$, and $\gamma$ satisfy the equations, then $\alpha+\pi$, $\beta$, and $\gamma$ would also satisfy the three tangent equations, but clearly $(\alpha+\pi)+\beta\ne\gamma$ in such a case. – boink Jun 15 at 19:19
• Thanks! :) ${}$ – Rajdeep Sindhu Jun 15 at 19:21

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
$$C= \dfrac{A+B}{1-AB},$$
• I don't know why, but, there seems to be some misconception here. Let's say that $\alpha + \beta$ is $\phi$ and $\gamma$ is another angle, let's assume that $\phi = \pi + \gamma$. So, $\tan (\alpha + \beta) = \tan\phi = \tan(\pi+\gamma) = \tan\gamma$, but $\alpha + \beta \neq \gamma$. This is essentially what my question was, shouldn't a condition be provided so as to imply the equality of $(\alpha + \beta)$ and $\gamma$ given that their outputs for the tangent function are equal. I was not referring to any condition within the identity. Did I miss something? – Rajdeep Sindhu Jun 15 at 20:17