I assume that you mean #cos(arctan(3/4))#, where #arctan(x)# is the inverse function of #tan(x)#.
(Sometimes #arctan(x)# as written as #tan^-1(x)#, but personally I find it confusing as it could be possibly misunderstood as #1/tan(x)# instead.)
We need to use the following identities:
#cos(x)=1/sec(x)# {Identity 1}
#tan^2(x)+1=sec^2(x)#, or #sec(x)=sqrt(tan^2(x)+1)# {Identity 2}
With these in mind, we can find #cos(arctan(3/4))# easily.
#\ \ \ \ \ \ \ cos(arctan(3/4))#
#=1/sec(arctan(3/4))# {Using Identity 1}
#=1/sqrt(tan(arctan(3/4))^2+1)# {Using Identity 2}
#=1/sqrt((3/4)^2+1)# {By definition of #arctan(x)#}
#=4/5#
