Question

How do you find tan (x+y) if tan x=5/4 and sec y=2?

Answer 1

`tan(x+y)=(5+4sqrt3)/(4-5sqrt3),` or,`=(5-4sqrt3)/(4+5sqrt3)`

Explanation:

Given that `secy=2`, we use the Identity `: sec^2y=1+tan^2y` to get, `tany=+-sqrt(sec^2y-1) = +-sqrt(4-1)=+-sqrt3.`

Now, `tan(x+y)=(tanx+tany)/(1-tanx*tany)=(5/4+-sqrt3)/(1-5/4*(+-sqrt3))`

`=(5+-4sqrt3)/(4-(+-5sqrt3)`

Thus, `tan(x+y)=(5+4sqrt3)/(4-5sqrt3),` or,`=(5-4sqrt3)/(4+5sqrt3)`