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Question:

Find $x$ if $\arctan(x+3)-\arctan(x-3)=\arctan(3/4)$

My attempt:

I know the formula: $$\arctan x-\arctan y=\arctan(\frac{x-y}{1+xy})$$ for both $x,y>0$

If both $x,y$ are NOT greater than $0$ then we'll need to form a second case in the above formula.

I solved assuming both $x+3$ and $x-3$ are greater than zero and got a quadratic in the end solving which I got $x=\pm4$.

I dutifully rejected $x=-4$ as it invalidated my assumption. However on putting in the above equation I found that it DOES satisfy the equation.

I do not wish to form cases for positive/negative. I wish to know if there is a simpler and direct way to solve such a question. Thanks!

Community
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Gaurang Tandon
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1 Answers1

2

Your answer $\{4,-4\}$ is right.

A full solution:

Since $\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$ and $\tan\arctan{x}=x$ for all real $x$, we obtain $$\tan\left(\arctan(x+3)-\arctan(x-3)\right)=\tan\arctan\frac{3}{4}$$ or $$\frac{x+3-(x-3)}{1+(x+3)(x-3)}=\frac{3}{4}$$ or $$x^2=16,$$ which gives $x=4$ or $x=-4$.

The checking of these numbers gets that indeed, they are roots and we are done!

Michael Rozenberg
  • 194,933
  • Thanks but when I'd proved the formula I had assumed that both $x,y>0$. Do you have some other proof where you can show this formula holds for all reals $x$ and $y$? – Gaurang Tandon Apr 21 '17 at 12:15
  • Yes, of course! Just use $\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$ – Michael Rozenberg Apr 21 '17 at 12:56
  • Uh, that's not a proof. Can you please show some steps? I'm willing to offer you a bounty if you can add the complete proof in your answer. – Gaurang Tandon Apr 22 '17 at 11:05
  • @Gaurang Tandon See now. – Michael Rozenberg Apr 22 '17 at 11:24
  • Sorry sir but your solution doesn't work always. For example, if I solve $\tan^{-1}(2x)+\tan^{-1}(3x)=3\pi/4$ using this method I'll get two solutions of which one value would be an extraneous value. – Gaurang Tandon Apr 24 '17 at 11:29
  • @Gaurang Tandon We need to check the roots, which we'll get. $x=1$ only gives a root and $-\frac{1}{6}$ is not suitable. In your starting equation we see that two roots are suitable. – Michael Rozenberg Apr 24 '17 at 12:48
  • @Gaurang Tandon Yes, you are right! In my solution I must add checking, otherwise my solution is not full. Thank you! I fixed my post. – Michael Rozenberg Apr 24 '17 at 12:59
  • Sir, I'd offer you a bounty, but please prove that the formula I've used in the topmost question is valid for all real $x,y$. The way you're doing it is taking tan on both sides and then reputting the values obtained back into the original equation - which is very long (and not simple or direct). Can you please prove that the formula I've used in the topmost question is valid for all real $x,y$. Thanks! – Gaurang Tandon Apr 24 '17 at 14:16
  • @Gaurang Tandon OK. Your formula is not true for all reals variables: $x=-1$ and $y=2$. It was my mistake. But my solution of your equation is true. – Michael Rozenberg Apr 24 '17 at 14:30