Euler's Totient function for semiprime numbers
You can get the 2nd equation from the first with some simple algebra.

$$(p-1)(q-1)$$
$$pq-p-q+1$$
$$n-p-q+1$$

I'm not really sure what you are asking though. Are you asking if computing the second is faster than the first?

I suppose you could argue that you have to do the $pq$ multiplication anyways, so why not reuse it. If that is your argument, then doing the calculation with the 2nd equation you gave for $\varphi$, then you get the multiplication for free and only have to do 2 subtractions and an addition. Using the first equation, you have to do two subtractions (although only a minus 1) and one multiplication. Multiplication is much more expensive, so the 2nd equation would be more efficient.