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Solve this equation : $\tan^{-1} \frac{x+1}{x-1} + \tan^{-1} \frac{x-1}{x} = \tan^{-1} (-7)$

This was an exam question, my try was as follows:

$$ \tan^{-1} \frac{x+1}{x-1} = \tan^{-1} (-7) - \tan^{-1} \frac{x-1}{x} $$

Now, assuming that $x = \tan x $ and substituting that in the above equation so that the equation changes and the $\tan^{-1} \frac{x-1}{x}$ vanishes, but, there is also one tan inverse function, so how to remove it?

Thanks :)

sky-fi
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1 Answers1

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Take tangents on the both sides, you'll get:

$$ \dfrac{\tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] + \tan \left [ \tan^{-1} \frac{x-1}{x} \right]}{ 1 - \tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] \tan \left [ \tan^{-1} \frac{x-1}{x} \right]} = -7 $$ $$ \frac{2x^2 - x + 1}{1-x} = -7 $$

Upon solving that equation, you'll get $x=2 \blacksquare$

PS: I used Wolframalpha to solve that equation, you may manually do it.

Note: I changed the question a bit, because I think that your question title is a bit incorrect, this question is example 6 of SL Loney's Inverse Trig Function

bantai_
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    Sorry, please comment here before you downvote my answer, I've changed the question a bit. – bantai_ Mar 19 '20 at 06:52
  • @Aditya Okay sir, Since the question was about Trigonometry, I've tried to show the important trigonometry steps, rest algebra is quite elementary, so, I guess it'd be fine to skip... Thanks for the suggestion, would keep that in mind in my future answers :) – bantai_ Mar 19 '20 at 06:58
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    In fact, the equation has no solutions. The solution $x=2$ needs to be checked because the two sides of the equation might differ by a multiple of $\pi$. That's the case here. – bjorn93 Mar 19 '20 at 07:50