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**arze** Solve the equation:

$\displaystyle \arcsin(\frac{x}{x-1}) +2 \arctan(\frac{1}{x+1})= \frac{\pi}{2}$

My attempt:

let $\displaystyle \theta = \arcsin(\frac{x}{x-1})$......1

and $\displaystyle \phi =\arctan(\frac{1}{x+1})$.......2

from these:

$\displaystyle \sin\theta=\frac{x}{x-1}$

$\displaystyle \tan\phi=\frac{1}{x+1}$

substitute 1 and 2 into the equation:

$\displaystyle \theta + 2 \phi = \frac{\pi}{2}$ OK up to here - see below.

$\displaystyle \sin (\theta+2\phi) =\sin \frac{\pi}{2}$

$\displaystyle \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1$

substitute $\displaystyle \sin\theta$ and use the identity:

$\displaystyle \sin2\phi=\frac{2t}{1+t^2}$ and $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$ where $\displaystyle t=\tan\phi$

$\displaystyle (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1$

$\displaystyle (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x +1})^2})=1$

my problem is the $\displaystyle \cos\theta$, I tried using pythagoras' theorem to find the the length of the adjacent side the then use that to determine the cosine, but I can't work it out.

thanks for any help.