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:

Using
the following formulas and
x*cos y - sin x* sin y

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

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

(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))

4/5)/(0 - sqrt(9/25))

(4/5)/(-3/5)

-4/3

Hence, evaluating the tangent of the
sum of inverse trigonometric functions yields