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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!

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      $\begingroup$ 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. $\endgroup$ – boink Jun 15 at 19:19
    • $\begingroup$ Thanks! :) ${}$ $\endgroup$ – Rajdeep Sindhu Jun 15 at 19:21
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    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.

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    • $\begingroup$ 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? $\endgroup$ – Rajdeep Sindhu Jun 15 at 20:17

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