Can the product $AB$ be computed using only $+, -,$ and reciprocal operators? Can the product of $A, B$ be computed using only $+, -,$ and reciprocal operators using a calculator? You can use calculator's memory function (multiply and divide are broken though).
Additional: I should have mentioned earlier, in addition to the 3 operators, the numberpad of the calculator can be used so yes 1 can be used.
 A: user2468's solution ends up requiring quite a lot of operations -- three reciprocals to get $A^2$, then the same to get $B^2$, then another three to get $A^2+B^2+2AB-A-B$.  Then to get rid of that pesky factor of 2 you need another three reciprocals, for a total of 12 reciprocal operations (and 31 binary operations).
It's possible to do better by employing some tricks:


*

*Instead of $(A+B)^2 - A^2 - B^2$, which involves three terms, use $(A+B)^2 - (A-B)^2$, which involves only two terms.

*Instead of needing to divide by a constant near the end of the process, which involves a lot more reciprocals, identify where in the formula you actually introduced the constant, and scale your other constants to match.


Using those tricks, I was able to reduce this down to 6 reciprocal operations and 11 binary operations: 
$$
\frac{1}{\frac{1}{A+B-2} - \frac{1}{A+B+2}}-\frac{1}{\frac{1}{A-B-2} - \frac{1}{A-B+2}} = AB
$$
I suspect this is probably the most beautiful solution to this challenge.
A: Edit: previous answer was wrong. Posted new answer. Hopefully right this time


*

*We can compute and store $A^2$ using
$$ \frac{1}{A} - \frac{1}{A+1} = \frac{1}{A^2 + A} $$
We can extract $A^2$ using only $+, -, ^{-1}.$ Similarly we can compute and store $B^2.$

*Then
$$\frac{1}{A+B-1} - \frac{1}{A+B}  = \frac{1}{(A+B)(A+B-1)} = \frac{1}{A^2 + B^2 + 2AB - A - B} $$
where we can extract $2AB,$ again, using only $+, -, ^{-1}$ and the values for $A^2, B^2$ we computed in step $1$ above.
Thanks to joriki, now to get $AB$ from $2AB$, add $\frac{1}{2AB} + \frac{1}{2AB},$ and take the reciprocal.
A: J.D. and N. S. have shown how to do it if a constant $1$ is allowed. Here's a proof that it can't be done if only $A$ and $B$ can be entered, no constants.
We can show by structural induction that all expressions we can generate change sign if both $A$ and $B$ change sign.
Base case: The two atomic expressions $A$ and $B$ change sign when both $A$ and $B$ change sign. 
Induction step: $x+y$ changes sign when both $x$ and $y$ change sign, $x-y$ changes sign when both $x$ and $y$ change sign, and $x^{-1}$ changes sign when $x$ changes sign.
Since $AB$ doesn't change sign when both $A$ and $B$ change sign, it follows that it can't be generated from $A$ and $B$ using only these operations.
A: We have $$\frac{A^2}4=\frac1{\dfrac1{A}-\dfrac1{A+4}}-A\tag1$$
and
$$AB=\frac14\,(A+B)^2-\frac14\,(A-B)^2\tag2.$$
