- #1

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## Homework Statement

[itex]\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}[/itex]

## Homework Equations

## The Attempt at a Solution

if [itex]\ arcsin\frac{x}{(x-1)} = α [/itex]

then [itex] sin α = \frac{x}{(x-1)}[/itex]

if [itex]2 arctan\frac{1}{x+1} = β[/itex]

then [itex]tan(\frac{β}{2}) = \frac{1}{(x+1)}[/itex]

[itex]tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}[/itex]

[itex]tan(β) = \frac{2(x+1)}{x(x+2)}[/itex]

[itex]α + β = \frac{π}{2}[/itex]

[itex]sin(α + β) = 1[/itex]

[itex]sin(α + β) = sinαcosβ + cosαsinβ[/itex]

[itex]cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}[/itex] by pythagoras' theorem

[itex](\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1[/itex]

[itex]\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1[/itex]

[itex]x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})[/itex]

At this point I figure I half made an error because I can't see a fruitful direction in which to progress and it feels a bit messy compared to the other problems in the exercise.