Answer:
[tex]\huge\boxed{g(a-1)=\dfrac{2-a}{a}=\dfrac{2}{a}-1}[/tex]
Step-by-step explanation:
[tex]g(t)=\dfrac{1-t}{1+t}\\\\g(a-1)-\text{substitute}\ t=a-1\ \text{to}\ g(t):\\\\g(a-1)=\dfrac{1-(a-1)}{1+(a-1)}=\dfrac{1-a-(-1)}{1+a+(-1)}=\dfrac{1-a+1}{1+a-1}=\dfrac{(1+1)-a}{(1-1)+a}=\dfrac{2-a}{a}[/tex]
