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Question:

Let there be an angle $\alpha$, such that $\tan \alpha = x$. Now we need to prove that $x \in \mathbb{R}$ and $\alpha \in (-90,90)$, meaning we have to prove that $x$ can take all real values without using graphs or any kind of graphs of prior knowledge and using the olympiad way of approach (i.e no calculus, limits ...)

My thoughts: Actually, I have tried thinking of how to solve this, but I cannot proceed even an inch further. I assumed $\tan \alpha \neq x$ and tried to contradict, but I'm not getting any intuition on how to further procced.

Any ideas?

Arbuja
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Name not Allowed
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    Maybe you mean $\alpha$ can't be a multiple of $90?$ $x$ can be any real number. Also $\alpha$ can be anything other than an odd multiple of $90.$ – Thomas Andrews Apr 11 '25 at 18:14
  • @ThomasAndrews Thanks for the correction! – Name not Allowed Apr 11 '25 at 18:38
  • You can do it if you define the right triangle, @jjagmath – Thomas Andrews Apr 11 '25 at 19:00
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    So you just want to draw a right triangle with legs $1$ and $x$ (and hypotenuse $\sqrt{x^2+1}$) and say you're done? – Ted Shifrin Apr 11 '25 at 19:31
  • Hint: Use the definition of $\tan x$ (two functions!), investigate the behavior of each function (one will increase, the other will decrease), and observe the behavior of the function near $90$ and $-90$. – bjcolby15 Apr 11 '25 at 19:41
  • But actually proving it, Can you please write a solution for that! – Name not Allowed Apr 11 '25 at 19:55
  • And I also want to know why some people have voted -1 and close votes? – Name not Allowed Apr 11 '25 at 19:56
  • You should note that $\tan(\alpha)$ is equal to the ratio $\frac{\sin(\alpha)}{\cos(\alpha)}$. As the angle $\alpha$ approaches $90^\circ$, the ratio $\frac{\sin(\alpha)}{\cos(\alpha)}$ approaches $\frac10$, which is $+\infty$. And as $\alpha$ approaches $-90^\circ$, the ratio approaches $\frac{-1}{0}$, which is $-\infty$. This shows that $\tan(\alpha)$ can cover all real numbers when $\alpha$ ranges from $-90^\circ$ to $90^\circ$. I’m not sure if this approach is allowed since you said we can’t use limits, but I just wanted to mention it. – VV_721 Apr 12 '25 at 00:47
  • @VV_721 that's a good thought, only problem is to make this a statement properly just like a formal proof – Name not Allowed Apr 12 '25 at 04:05
  • @NamenotAllowed "Here's my problem - solve it for me" without any effort on the OP is often a good reason for readers to downvote and request posts be closed. If you gave us points where you don't understand what's going on (i.e. "I was stuck here, am I on the right path?") then we can help. And, readers run the risk of being downvoted for giving too much help in an answer, so we give hints. In fact, I'll give you a few once I've finished this post (VV_721 may have given you some already). – bjcolby15 Apr 12 '25 at 10:07
  • @bjcolby15 Thanks! – Name not Allowed Apr 12 '25 at 11:05
  • @NamenotAllowed Anytime! Even if you put something in the comments section, that may trigger a little nudge from the Stack Exchange hivemind. Hopefully what I put down is helpful. – bjcolby15 Apr 12 '25 at 11:44

1 Answers1

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Hints:

  1. The definition of $\tan x$ is $\dfrac {\sin x}{\cos x}$. As $\sin x$ approaches $90°$, what happens to $\cos x$? What happens if it approaches $-90°$?

  2. When you reach $90°$ or $-90°$ exactly, what happens to $\sin x$ and $\cos x$?

If you're still stuck, see this post (this was my answer back in 2018)...Finding the values of possible values of tan x

bjcolby15
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