Find the Exact Value

$\cos^{-1}(1) = 0\\ \tan\left(\sin^{-1}\left(\dfrac{4}{5}\right)+\cos^{-1}(1)\right)\\ = \tan\left(\sin^{-1}\left(\dfrac{4}{5}\right)\right)\\ = \dfrac{4}{\sqrt{5^2-4^2}}\\ =\dfrac{4}{3}$

$\tan\left(\cos^{-1}\left(\dfrac{4}5{}\right)+\sin^{-1}\left(1\right)\right)\\ =\tan\left(\dfrac{\pi}{2}+\cos^{-1}\left(\dfrac{4}5\right)\right)\\ =\dfrac{1}{\tan\left(\cos^{-1}\left(\dfrac{4}5\right)\right)}\\ =\dfrac 1{\dfrac{\sqrt{5^2-4^2}}4}\\ =\dfrac{4}{3}$

tan ( arccos(4/5) + arcsin(1) ) =

tan (arcos(4/5) + pi/2)

Let    arccos(4/5)   =  θ           so  cos θ = 4/5       and sin θ = 3/5

So we have

tan ( θ + pi/2)  =

sin ( θ + pi/2)               sinθ cos pi/2   +  sin pi/2 * cos θ                 cos θ

___________   =        ___________________________   =    __________ =

cos (θ + pi/2)               cos θ cos pi/2  -   sin θ sin pi/2                  - sin θ  

(4/5)             - 4

____    =     ___

-(3/5)             3

I believe this is what hectictar was pointing out.......