Answer
Solution set: $\{-\sqrt{3},\sqrt{3}\}$
Work Step by Step
$f(a)=g(a)+1$
$\displaystyle \frac{a+2}{a+3}=\frac{a+1}{a^{2}+2a-3}$ ... factor all denominators
$a^{2}+2a-3=(a-1)(a+3)$
$\displaystyle \frac{a+2}{a+3}=\frac{a+1}{(a-1)(a+3) } \quad $... LCD=$(a-1)(a+3)$
Exclude solutions which yield 0 in denominators:
$a\not\in\{-3,1\}\qquad (*)$
Multiply the equation with the LCD=$(a-1)(a+3)$
$(a+2)(a-1)=a+1$
$ a^{2}+a-2=a+1\qquad$ ... add $-a+2$
$a^{2}=3$
$ a=\pm\sqrt{3}\qquad$ .... both satisfy (*).
Solution set: $\{-\sqrt{3},\sqrt{3}\}$
