Question about a trigonometry proof? I just want to ask how can you prove that 2α is twice the value of α in the following figure that depicts a proof of an arctangent identity (and likewise, for β as well).

 A: You should realize that the upper horizontal line (the line $y=1$) is tangent to the circle that is drawn. We use the following result:

The angle between a circle's tangent and a cord of that circle is equal to half of the central angle induced by that cord. (See image)


(To see this you must know some basic geometrical theorems regarding circles.)
A: let $$a = \tan^{-1}\left(\frac{1-y}x\right), b = \tan^{-1}\left(\frac{1-x}y\right) .$$  then 
$$\begin{align}\tan(a+b) &= \frac{\tan a + \tan b}{1- \tan a \tan b} \\
&= \frac{\frac{1-y}x+\frac{1-x}y}{1-\frac{1-y}x \frac{1-x}y} \\
&= \frac{y+x-y^2-x^2}{xy-(1-x-y+xy)} \\
&=\frac{(x+y)(1-x-y)}{-(1-x-y)} \\
&= -x-y\end{align}$$
don't you need $x+y = -1$  for $a+b = \pi/4?$
A: Denote $\theta$ your angle (until we prove it is truly $\alpha$). Consider the upper isosceles triangle with vertex at the origin. Its two equal angles are equal to $\dfrac\pi2-\theta$, hence:
$$2\Bigl(\dfrac\pi2-\theta\Bigr)=\pi -2\alpha \Rightarrow \dfrac\pi2-\theta=\dfrac\pi2-\alpha. $$
A: WLOG let $x=\cos2u,y=\sin2u$
$\dfrac{1-x}y=\dfrac{1-\cos2u}{\sin2u}=\dfrac{2\sin^2u}{2\sin u\cos u}=\tan u$
$\dfrac{1-y}x=\dfrac{1-\sin2u}{\cos2u}=\dfrac{(\cos u-\sin u)^2}{\cos^2u-\sin^2u}=\dfrac{\cos u-\sin u}{\cos u+\sin u}=\dfrac{1-\tan u}{1+\tan u}=\tan\left(\dfrac\pi4-u\right)$
A: 
We assume $AC$ is the tangent line to the circle at $A$, which means $\widehat{OAC}=90$. Showing the tangent chord angle is half the central angle (i.e, your $\alpha$ and $2\alpha$ angles) is equivalent to showing
\begin{equation} \widehat{BAC} = \frac{1}{2} \widehat{AOB}. \end{equation}
Since $OA=OB$ (the radius of a circle) and the sum of all angles in $\Delta OAB$ (isosceles triangle) is $180$, we have 
\begin{equation} \widehat{OAB}=\widehat{OBA} = \frac{1}{2}(180 - \widehat{AOB}). \end{equation}
Using this equality and the fact that $\widehat{OAC}=90$, we obtain
\begin{equation}
\widehat{BAC} = 90 - \widehat{OAB} = 90 - \frac{1}{2}(180 - \widehat{AOB}) = \frac{1}{2} \widehat{AOB}
\end{equation}
as desired.
Now the rest is exactly what you wrote.
