Originally Posted by

**mj.alawami** **Question**:

if $\displaystyle tanx=\frac{7}{3} $ and $\displaystyle 0<x<\frac{\pi}{2} $;

$\displaystyle tany=0$ and $\displaystyle 0<x<\frac{\pi}{2} $***

tan(y) = 0 ?

then y = 0, pi, or 2pi ... which one?

Find

a)tan(x+y)

b)tan(x-y)

c) the quadrant containing (x-y)

d)the quadrant containing (x+y)

Attempt:

Using the formula

$\displaystyle tan(x -/+ y) \frac{tanx +/- tany}{1 +/- tanx*tany} $

formula correction ...

$\displaystyle \tan(x \pm y) = \frac{\tan{x} \pm \tan{y}}{1 \mp \tan{x} \cdot \tan{y}}$