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!

  • $\endgroup$
    • 2
      $\begingroup$ 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

    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.

    | cite | improve this answer | |
    • $\begingroup$ 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

    Your Answer

    By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

    Not the answer you're looking for? Browse other questions tagged or ask your own question.