yasusar0

2022-07-18

Solving inverse trig question without forming cases
Question:
Find $x$ if $\mathrm{arctan}\left(x+3\right)-\mathrm{arctan}\left(x-3\right)=\mathrm{arctan}\left(3/4\right)$
My attempt:
I know the formula:
$\mathrm{arctan}x-\mathrm{arctan}y=\mathrm{arctan}\left(\frac{x-y}{1+xy}\right)$
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 $x=±4$
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!

Vartavk

Your answer $\left\{4,-4\right\}$ is right.
A full solution:
Since $\mathrm{tan}\left(\alpha -\beta \right)=\frac{\mathrm{tan}\alpha -\mathrm{tan}\beta }{1+\mathrm{tan}\alpha \mathrm{tan}\beta }$ and $\mathrm{tan}\mathrm{arctan}x=x$ for all real $x$, we obtain
$\mathrm{tan}\left(\mathrm{arctan}\left(x+3\right)-\mathrm{arctan}\left(x-3\right)\right)=\mathrm{tan}\mathrm{arctan}\frac{3}{4}$
or
$\frac{x+3-\left(x-3\right)}{1+\left(x+3\right)\left(x-3\right)}=\frac{3}{4}$
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!

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