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

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

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

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

why it equals
[math]\frac{\frac{-2L}{3}}{R}=-1[/math]
 
nhrock3 said:
-\frac{\pi}{4}=arctan(\frac{\frac{-2L}{3}}{R})\\

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

?
Take the tangent of both sides.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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