find f'(pi/4) of f(x) = [ln (cos x)]^2

I know f'(x) = -2 tan x ln (cos x)

two questions for f'(pi/4)

I know the answer is ln 2

so I know I compute -2 tan pi/4 ln (cos pi/4)

first question is tan pi/4 * cos pi/4 become 1/sqrt2. is there an easy way of remember such things?

next the problem then becomes -2 ln (2^-1/2)

which becomes ln 2. I am missing how it becomes ln 2. ??