If $((p-1)*(q-1) -1)$ divisible by $e$ ($e$ is odd number) , then $\text{gcd}(e,(p-1)*(q-1)) = 1$ If $((p-1)(q-1) -1)$ divisible by $e$ ($e$ is odd number) , then $\text{gcd}(e,(p-1)(q-1)) = 1$. ($p,q$ are prime numbers ) Is this true, if yes why, if not why not ?
Is this true?

I'd generalize it, and replace $(p-1)(q-1)$ with $X$; that is, if $X-1$ is divisible by $e$, that is, if $X-1 = k \times e$ for some integer $k$, then $\gcd(e, X) = 1$. Is this true?  If it is, then your original statement is also true (because if it holds for all $X, e$, it also holds for all $X$ that is of the form $(p-1)(q-1)$ and for all odd $e$.).
As a further hint, we can also observe that $\gcd(e, X) = \gcd(e, X \bmod e)$ (that's the central observation that makes the Euclidean method for evaluating $\gcd$ work).
With that, it should be fairly straight-forward to proceed...