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!
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!