Find the value of `cos(tan^(-1)(2)+tan^(-1)(3))`?

First use the cosine angle-addition formula:

`cos(a+b)=cos(a)cos(b)-sin(a)sin(b)`

Then the original expression equals:

`=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))`

Note that if `theta=tan^-1(2)`, then `tan(theta)=2`. That is, a right triangle with angle `theta` has `tan(theta)=2`, which is the triangle with the leg opposite `theta` being `2` and the leg adjacent to `theta` being `1`. The Pythagorean theorem tells us that the hypotenuse of this triangle is `sqrt5`.

So, when `theta=tan^-1(2)`, we see that:

`cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5`

`sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5`

Now let angle `phi` be defined by `phi=tan^-1(3)`, such that `tan(phi)=3`. This is the right triangle with:

`{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}`

Then:

`cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10`

`sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10`

Plugging these into the expression from earlier we get:

`=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)`

Note that `sqrt5(sqrt10)=sqrt50=5sqrt2`:

`=(1-6)/(5sqrt2)=-1/sqrt2`

Let `tan^(-1)(2)=alpha` and `tan^(-1)(3)=beta`

Therefore `tanalpha=2` and `tanbeta=3`

As `tan(pi/4)=1`, `alpha` and `beta` lie between `pi/4` and `pi/2`

and hence `(alpha+beta)` lies in Q2.

and `tan(alpha+beta)=(tanalpha+tanbeta)/((1-tanalphatanbeta)`

= `(2+3)/(1-2xx3)=5/(-5)=-1` and `(alpha+beta)=(3pi)/4`

Hence, `cos(tan^(-1)(2)+tan^(-1)(3))=cos((3pi)/4)=-1/sqrt2`