Q:

Prove $P(A|B^c)=P(A)$

Given that $A$ and $B$ are independent, Prove $P(A|B^c)=P(A)$

$P(A|B^c)=P(A\cap B^c)/P(B^c)=P(A)$ if $A$ is independent to $B^c$, which is not given.
Am I wrong?

A:

Well, then you will need to show that $A$ and $B^{c}$ are independent if $A$ and $B$ are.
So we need to show $P(A \cap B^{c}) = P(A)P(B^{c})$.
Let's do this using other equations we already know that involve some of the above terms:

$P(A) = P(A \cap B) + P(A \cap B^{c})$, right?
Also, since $A$ and $B$ are independent, $P(A \cap B) = P(A)P(B)$.

So substituting 2. into 1. gives us $P(A) = P(A)P(B) + P(A \cap B^{c})$, or rearranged: $$P(A) - P(A)P(B) = P(A \cap B^{c}).$$
Now, if only the left hand side (LHS) equaled $P(A)P(B^{c})$, we'd be done!  
But wait:
$$\text{LHS} = P(A) - P(A)P(B) = P(A)(1 - P(B))$$ and since $1 - P(B) = P(B^{c})$, we do get that the left hand side equals $P(A)P(B^{c})$.
So, $P(A \cap B^{c}) = P(A)P(B^{c})$, which means $A$ and $B^{c}$ are independent if $A$ and $B$ are.

