tan(cos^-1(4/5)+sin^-1(1)) I need to solve without using a calculator a step by step process I need to use tan(a+B)= Sin (a+B) / cos (a+B)...

You need to use the fact that the tangent function is a
rational function such that: `tan x = sin x/cos x`

Using
the following formulas `sin (x+y) = sin x*cos y + sin y*cos x ` and `cos(x+y) = cos
x*cos y - sin x* sin y`  yields:

`tan
(cos^-1(4/5)+sin^-1(1)) = (sin (cos^-1(4/5)+sin^-1(1)))/(cos
(cos^-1(4/5)+sin^-1(1)))`

`tan (cos^-1(4/5)+sin^-1(1)) =
(sin (cos^-1(4/5))*cos(sin^-1(1)) +
sin(sin^-1(1))*cos(cos^-1(4/5)))/(cos(cos^-1(4/5))*cos(sin^-1(1)) -
sin(sin^-1(1))*sin(cos^-1(4/5)))`

Use `sin(sin^-1 a) = a` 
and `sin(cos ^-1 a) = sqrt(1 - a^2),`

` cos(cos^-1 a) = a
and cos(sin ^-1 a) = sqrt(1 - a^2).`

`tan
(cos^-1(4/5)+sin^-1(1)) = (sqrt((1 - 16/25)(1 - 1)) + 1*(4/5))/((4/5)*sqrt(1-1) -
1*sqrt(1 - 16/25))`

`tan (cos^-1(4/5)+sin^-1(1)) = (0 +
4/5)/(0 - sqrt(9/25))`

`tan (cos^-1(4/5)+sin^-1(1)) =
(4/5)/(-3/5)`

`tan (cos^-1(4/5)+sin^-1(1)) =
-4/3`

Hence, evaluating the tangent of the
sum of inverse trigonometric functions yields`tan (cos^-1(4/5)+sin^-1(1)) = -4/3.`