$ T(n)=T(√n)+loglog(n) $
Let $ n=2^k$
recurrence equation will be
$T(2^k)=T(2^{k/2})+loglog(2^k)$
Let $T(2^k)=S(k)$
So $S(k)=S(k/2)+log (k)$
By masters Theorem
$ a=1 , b=2, f(k)= log (k) $
2nd Case Master theorem not possible. It's 3rd case because here $ f(k)= log (k )$ is greater than $k^{log _{b}^{a}}=k^{log_{2}^{1}}=k^0$ but not polynomial time.
So,
$S(k)$ = $\Theta$$(k^0 * log( k)^{1+1})$=$\Theta$ $(log^{2} k)$
Replacing all values with T(n) and its equivalent and k= log n.
So $T(n)$ = $\Theta(log^{2}log(n))$. //k=log(n).