Why Does $\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})$ Equal -1?

[nhrock3]

$$-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\$$

why it equals
$$\frac{\frac{-2L}{3}}{R}=-1$$

?

 

[dacruick]

[math]-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\[/math]

why it equals
[math]\frac{\frac{-2L}{3}}{R}=-1[/math]

 

[Mark44]

$$-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\$$

why it equals
$$\frac{\frac{-2L}{3}}{R}=-1$$

?

Take the tangent of both sides.