Although this was in the "Differentiating Trigonometric Functions" section, it looks like you just want to evaluate or simplify what you provided.
#y=3sec^-1(sqrt(2))-4csc^-1(sqrt(2))+5cot^-1(2)#
FIRST TERM: #sec^-1(sqrt(2))#
If #x=sec^-1(sqrt(2))#, then #sec(x)=sqrt(2)# or #1/cos(x)=sqrt(2)#.
This is the same as saying
#cos(x)=1/sqrt(2)=sqrt(2)/2#, which happens when #x=pi/4#
SECOND TERM: #csc^-1(sqrt(2))#
If #x=csc^-1(sqrt(2))#, then #csc(x)=sqrt(2)#
Identically, #1/sin(x)=sqrt(2)# or #sin(x)=1/sqrt(2)=sqrt(2)/2#, which also happens at #x=pi/4#
THIRD TERM: #cot^-1(2)#
If #x=cot^-1(2)#, then #cot(x)=2# or #1/tan(x)=2# or #tan(x)=1/2#. Because the function #y=tan(x)# repeats periodically (see graph), you actually get an infinite number of possible #x# values. So you can't really say what #cot^-1(2)# is.
graph{y=tan(x) [-20,20,-0.5,1]}
Most calculators assume you meant the middle one though and will give you an answer #cot^-1(2)~~0.4636#
ANSWER
Plugging those values in, you get
#y=3(pi/4)-4(pi/4)+5cot^-1(2)#
#y=5cot^-1(2)-pi/4#
A calculator will deceptively say #y~~1.5326#
