# Thread: Solving sum and difference identities

1. ## Solving sum and difference identities

Question:
if $tanx=\frac{7}{3}$ and $0;
$tany=0$ and $0
Find
a)tan(x+y)
b)tan(x-y)

Attempt:

Using the formula
$tan(x -/+ y) \frac{tanx +/- tany}{1 +/- tanx*tany}$
a) $\frac{7}{3}$
b) $\frac{7}{3}$

Is my attempt Correct?

Thank you

2. Originally Posted by mj.alawami
Question:
if $tanx=\frac{7}{3}$ and $0;
$tany=0$ and $0***

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

Find
a)tan(x+y)
b)tan(x-y)

Attempt:

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

formula correction ...

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

3. Originally Posted by skeeter
.
Ooh I am very sorry I misread the question properly

$tanx=\frac{7}{3}$ and it is in the first quadrant (0<x<[tex]\frac{\pi}{2});

$tany=\frac{3}{4}$ and it is in the first quadrant (0<y<[tex]\frac{\pi}{2})

Find
a) tan(x+y)
b)tan (x-y)

Attempt:
a) $\frac{19}{33}$
b) $\frac{-37}{9}$

4. Originally Posted by mj.alawami
Ooh I am very sorry I misread the question properly

$tanx=\frac{7}{3}$ and it is in the first quadrant (0<x<[tex]\frac{\pi}{2});

$tany=\frac{3}{4}$ and it is in the first quadrant (0<y<[tex]\frac{\pi}{2})

Find
a) tan(x+y)
b)tan (x-y)
a) $\frac{19}{33}$
b) $\frac{-37}{9}$
$\tan(x+y) = -\frac{37}{9}$ ... quad II
$\tan(x-y) = \frac{19}{33}$ ... quad I