How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$? How to compute $\frac{t}{t+1}$ to the form $1-\frac{1}{t+1}$? 
What else? Well.
Well can you use long division?
 A: Of course you can use long division to show that $$\dfrac t{t+1} = 1  - \frac{1}{t+1}$$
But a simpler route is to note that $t = t + 1 - 1$, which gives us $$\dfrac t{t+1} = \dfrac{(t+1) -1}{(t+1)} = \dfrac{(t+1)}{(t+1)} - \dfrac {1}{t+1} = 1 - \dfrac{1}{t+1}$$
A: You can't, since it is never true (for it to be true, we would require $t=t-2$).
You can however show that $$\frac{t}{t-1}=1+\frac{1}{t-1},$$ by observing that $t=t-1+1,$ and you can show that $$\frac{t}{t+1}=1-\frac{1}{t+1},$$ by observing that $t=t+1-1.$
A: Use method add something and take something.
$$\frac{t}{t-1}= \frac{t-1+1}{t-1}$$
Then splits equation on 2 parts.
$$\frac{t-1}{t-1}+ \frac{1}{t-1}  $$ 
Then reduce fraction and you get:  $$1+\frac{1}{t-1}$$
If this is true
$$\frac{t}{t-1}=  1-\frac{1}{t-1}$$
Than you get:
$$\frac{t}{t-1}=  \frac{t-1-1}{t-1}  $$ 
Than this equation  multiply with $(t-1)$
$$t=t-1-1 \Rightarrow t=t-2 \Rightarrow 0t=-2$$  
So we take  $t=t-2$
$$\frac{t-2}{t-2-1}= \frac{t-2-1+1}{t-2-1}$$
$$\frac{t-2-1}{t-2-1}+ \frac{1}{t-2-1}$$
We get solution which is not the same as we previously calculated
$$1 + \frac{1}{t-2-1}=1 + \frac{1}{t-3}$$
So
$$\frac{t}{t-1} \neq 1 -\frac{1}{t-1}$$

$$\frac{t}{t+1} = \frac{t+1-1}{t+1} = \frac{t+1}{t+1} -\frac{1}{t+1}= 1 -\frac{1}{t+1} $$
A: HUGE EDIT IN RESPONSE TO OP CORRECTION:
$$\dfrac{t}{t+1} = \dfrac{t + 1 - 1}{t+1} = \dfrac{t + 1}{t+1} - \dfrac{1}{t+1} = 1 - \dfrac{1}{t+1}$$
When in doubt, remember your identities: try adding a weird form of zero (in this case 1-1), try multiplying by a weird form of 1, and in more advanced math, use your trig identities.
