SOLUTION: I hope you can help me :) problem is.. Find (a) tan(x+y) (b) tan(x-y) through tan x = 7/3,0<x<1/2pi ; tan y= 3/4, 0<x<1/2pi. Thanks!

Algebra ->  Trigonometry-basics -> SOLUTION: I hope you can help me :) problem is.. Find (a) tan(x+y) (b) tan(x-y) through tan x = 7/3,0<x<1/2pi ; tan y= 3/4, 0<x<1/2pi. Thanks!      Log On


   

Question 511104: I hope you can help me :)
problem is..
Find (a) tan(x+y) (b) tan(x-y) through tan x = 7/3,0 Thanks!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find
(a) tan(x+y)
= [tan(x)+tan(y)]/[1-tan(x)*tan(y)]
---
= [7/3 + 3/4]/[1 - (7/3)(3/4)]
---
= [37/12] / [1 - (7/4)]
---
= (37/12)/(-3/4)
--
= (37/12)*(-4/3)
= -36/9
=========================
(b) tan(x-y)
= [tan(x)-tan(y)]/[1+tan(x)*tan(y)]
---
= [7/3 - 3/4]/[1 + (7/3)(3/4)]
---
= [18/12] / [1 + (7/4)]
---
= (3/2)/(11/4)
--
= (3/2)*(4/11)
----
= 6/11

---
tan x = 7/3, 0 ===============
Cheers,
Stan H.