# 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)...

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)

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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.` **

tan(a+b) = (sin(a)cos(b)+sin(b)cos(a))/(cos(a)cos(b)-sin(a)sin(b))

sin(cos^-1(4/5)) = sqrt(1-cos^2(cos^-1(4/5))) = sqrt(1-(4/5)^2) = sqrt(1-16/25) = sqrt(9/25) = 3/5

sin(sin^-1(1)) = 1

cos(cos^-1(4/5) = 4/5

cos(sin^-1(1)) = sqrt(1-sin^2(sin^-1(1))) = sqrt(1-1^2) = 0

so tan(a+b) = (3/5*0+4/5*1)/(4/5*0-3/5*1) = (4/5)/(-3/5) = -4/3