Question

How do you find the exact value `tan(x+y)` if `cotx=6/5,secy=3/2`?

Answer 1

If `y` is in `Q1`, `tan(x+y)=(5+3sqrt5)/(6-5sqrt5)` and
if `y` is in `Q4`, `tan(x+y)=(5-3sqrt5)/(6+5sqrt5)`

Explanation:

As `cotx=6/5`, `tanx=1/cotx=1/(6/5)=5/6`

and as `secy=3/2`, `tany=+-sqrt((3/2)^2-1)=+-sqrt(9/4-1)=+-sqrt(5/4)=+-sqrt5/2`

Observe that `tany=sqrt5/2`, when `y` is in `Q1` and `tany=-sqrt5/2`, when `y` is in `Q4`.

Hence there canj be two values of `tan(x+y)`

If `y` is in `Q1`, `tan(x+y)=(tanx+tany)/(1-tanxtany)=(5/6+sqrt5/2)/(1-5/6xxsqrt5/2)=(5+3sqrt5)/(6-5sqrt5)`

If `y` is in `Q4`, `tan(x+y)=(tanx+tany)/(1-tanxtany)=(5/6-sqrt5/2)/(1+5/6xxsqrt5/2)=(5-3sqrt5)/(6+5sqrt5)`