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

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
