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
