Hello, classicstrings!
How would I express $\displaystyle \tan(270^o-\theta)$ in terms of $\displaystyle \tan\theta$?
You're right . . . the compound-angle formula gives us $\displaystyle \tan270^o$
We must find a way around this dilemma . . .
Graph the angle . . . Code:
Y
| *Q
| /
|θ/
|/
- - - - + - - - - X
/|O
/θ|
/ |
/ |
P* |
Y'
$\displaystyle \angle XOP = 270^o - \theta\quad\Rightarrow\quad \angle POY' = \theta\quad\Rightarrow\quad \angle QOY = \theta$
Hence: .$\displaystyle \tan(270^o - \theta) \:=\:\tan\left(90^o - \theta\right) \:=\:\frac{1}{\tan\theta}$