Question

A= 2 arctan((1-x)÷(1+x))
B= arcsin((1-x^2)÷(1+x^2))
prove A+B=pie.

Answer 1

\[\begin{cases}
A=2\arctan\dfrac{1-x}{1+x}\\[1ex]
B=\arcsin\dfrac{1-x^2}{1+x^2}
\end{cases}\]
This *seems* to be a step in the right direction, but I'm not sure how to finish. First write
\[\begin{cases}
\tan\dfrac{A}{2}=\dfrac{1-x}{1+x}\\[1ex]
\sin B=\dfrac{1-x^2}{1+x^2}
\end{cases}\]
We then have
\[\begin{align*}
\sin B\tan\frac{A}{2}+\sin B\cot\frac{A}{2}&=\sin B\tan\frac{A}{2}\left(1+\cot^2\frac{A}{2}\right)\\[1ex]
&=\sin B\tan\frac{A}{2}\csc^2\frac{A}{2}\\[1ex]
&=\frac{\sin B}{\sin\dfrac{A}{2}\cos\dfrac{A}{2}}\\[1ex]
&=\frac{2\sin B}{\sin A}\end{align*}\]
Meanwhile,
\[\begin{align*}
\sin B\tan\frac{A}{2}+\sin B\cot\frac{A}{2}&=\frac{1-x^2}{1+x^2}\times\frac{1-x}{1+x}+\frac{1-x^2}{1+x^2}\times\frac{1+x}{1-x}\\[1ex]
&=\frac{(1-x)^2}{1+x^2}+\frac{(1+x)^2}{1+x^2}\\[1ex]
&=\frac{2+2x^2}{1+x^2}\\[1ex]
&=2
\end{align*}\]
So it turns out that \(\dfrac{2\sin B}{\sin A}=2\), or \(\sin B=\sin A\).

Answer 2

Sry dude i ask something and you answered something. Just see the question.

Answer 3

Is \(x\) restricted to belong to a certain range? Notice that if \(x=1\), you have \(A+B=0\).