1. ## Exact Values

How would I express tan(270-A) in terms of tanA?

Would I use the double angle formula for tan? But that gives tan270 which is undefined.?

2. Originally Posted by classicstrings
How would I express tan(270-A) in terms of tanA?

Would I use the double angle formula for tan? But that gives tan270 which is undefined.?
Hello,

use the property:

$\tan(270^\circ-A)=\frac{\sin(270^\circ-A)}{\cos(270^\circ-a)}$ $=\frac{\sin(270^\circ) \cdot \cos(A)-\cos(270^\circ) \cdot \sin(A)}{\cos(270^\circ) \cdot \cos(A)-\sin(270^\circ) \cdot \sin(A)}$

Now sin(270°) = -1 and cos(270°) = 0

Therefore you get:

$\tan(270^\circ-A)=\frac{-\cos(A)}{-\sin(A)}=\frac{1}{\tan(A)}$

EB

3. Thanks earboth!

4. Hello, classicstrings!

How would I express $\tan(270^o-\theta)$ in terms of $\tan\theta$?

You're right . . . the compound-angle formula gives us $\tan270^o$
We must find a way around this dilemma . . .

Graph the angle . . .
Code:
            Y
|   *Q
|  /
|θ/
|/
- - - - + - - - - X
/|O
/θ|
/  |
/   |
P*    |
Y'

$\angle XOP = 270^o - \theta\quad\Rightarrow\quad \angle POY' = \theta\quad\Rightarrow\quad \angle QOY = \theta$

Hence: . $\tan(270^o - \theta) \:=\:\tan\left(90^o - \theta\right) \:=\:\frac{1}{\tan\theta}$