First we expand by using formula for cosine of sum
#[cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)]#
#cos((tan^-1(2)))cos((tan^-1(3)))-sin((tan^-1(2)))sin((tan^-1(3)))#
We need to express #cos(x)# and #sin(x)# using #tan(x)# only.
We know that
#{(sin^2(x)+cos^2(x)=1),(sin(x)/cos(x)=tan(x)):}#
We can multiply second equation by cos^2 and square it
#{(sin^2(x)+cos^2(x)=1),(color(red)(sin^2(x)=tan^2(x)cos^2(x))):}#
Now we can solve it like linear equation with variables #sin^2(x)# and #cos^2(x)#
Substiture
#{(color(red)(tan^2(x)cos^2(x))+cos^2(x)=1larr),(sin^2(x)=tan^2(x)cos^2(x)rarr):}#
Factor out and divide first to get
#cos^2(x)=1/(tan^2(x)+1)#
and then
#sin^2(x)=tan^2(x)/(tan^2(x)+1)#
So
#sin(x)=tan(x)/sqrt(tan^2(x)+1)# and #cos(x)=1/sqrt(tan^2(x)+1)#
Plugging #tan^-1(y)# in to #cos(x)# gives us
#cos(tan^-1(y))=1/sqrt(tan^2(tan^-1(y))+1)#
#cos(tan^-1(y))=1/sqrt((cancel(tan)(cancel(tan^-1)(y)))^2+1)#
#cos(tan^-1(y))=1/sqrt(y^2+1)#
similarly
#sin(tan^-1(y))=y/sqrt(y^2+1)#
So the original expression becomes
#1/sqrt(2^2+1)*1/sqrt(3^2+1)-2/sqrt(2^2+1)*3/sqrt(3^2+1)#
#=1/(sqrt(5)*sqrt(10))-6/(sqrt(5)*sqrt(10))#
#=-5/(5*sqrt2)=-1/sqrt2#
