Question 1165381
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{\sin\theta}{\cos\theta}]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ +\ \sin^2\theta\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ 1\ -\ \cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\sqrt{1\,-\,\cos^2\theta}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{\pm\sqrt{1\,-\,\cos^2\theta}}{\cos\theta}]


But select the negative because *[tex \Large \theta\ \in\ \text{QIII}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{-\sqrt{1\,-\,\cos^2\theta}}{\cos\theta}]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>