## anonymous one year ago A= 2 arctan((1-x)÷(1+x)) B= arcsin((1-x^2)÷(1+x^2)) prove A+B=pie.

1. anonymous

$\begin{cases} A=2\arctan\dfrac{1-x}{1+x}\\[1ex] B=\arcsin\dfrac{1-x^2}{1+x^2} \end{cases}$ 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 $$\sin B=\sin A$$.

2. anonymous

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

3. anonymous

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

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