Solving for x in arcsin+arctan=π/2

[Appleton]

Homework Statement

$\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}$

Homework Equations

The Attempt at a Solution

if $\ arcsin\frac{x}{(x-1)} = α$

then $sin α = \frac{x}{(x-1)}$

if $2 arctan\frac{1}{x+1} = β$

then $tan(\frac{β}{2}) = \frac{1}{(x+1)}$

$tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}$

$tan(β) = \frac{2(x+1)}{x(x+2)}$

$α + β = \frac{π}{2}$

$sin(α + β) = 1$

$sin(α + β) = sinαcosβ + cosαsinβ$

$cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

$sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

$cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}$ by pythagoras' theorem

$(\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1$

$\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1$

$x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})$

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.

 

[eumyang]

I haven't figured the answer myself, but I have a couple of comments:

$\cos β = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

$\sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

The denominators here will simplify into an irreducible quadratic.

$\cos α = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}$ by pythagoras' theorem

TBH, I don't know if this will be helpful, but the numerator can be simplified a bit (although you won't lose the square root).

 

[verty]

Homework Statement

$\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}$

Homework Equations

The Attempt at a Solution

if $\ arcsin\frac{x}{(x-1)} = α$

then $sin α = \frac{x}{(x-1)}$

if $2 arctan\frac{1}{x+1} = β$

then $tan(\frac{β}{2}) = \frac{1}{(x+1)}$

$tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}$

$tan(β) = \frac{2(x+1)}{x(x+2)}$

$α + β = \frac{π}{2}$

$sin(α + β) = 1$

$sin(α + β) = sinαcosβ + cosαsinβ$

$cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

$sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}$ by pythagoras' theorem

$cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}$ by pythagoras' theorem

$(\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1$

$\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1$

$x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})$

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.

You were on the right track at this point:

$α + β = \frac{π}{2}$

but the next step was wrong. Can you think of something better to do here?

Hint: Don't make it any more complicated than it already is.

Spoiler

Use sin and cos

 

[Appleton]

Thanks for your comments. After staring at this for a good proportion of the day I noticed that α and β are complementary angles which enabled me to proceed in the following fashion:

$sinα = \frac{x}{x-1}$

$cosβ = \frac{x}{x-1}$ complementary angle

$cosα = \frac{-2x+1}{x-1}$ pythagoras' theorem

$sinβ = \frac{-2x+1}{x-1}$ complementary angle

$sin(α+β) = 1$

$sin(α+β) = (\frac{x}{x-1})^2 + (\frac{-2x+1}{x-1})^2 = 1$

$\frac{5x^2-4x+1}{x-1} = 1$

$5x^2-4x+1 = x-1$

$x^2-x = 0$

$x(x-1) = 0$

$x = 0$ or $x = 1$

but $asin \frac{1}{1-1}$ makes no sense

so $x = 0$

However the process I have deployed here seems no less complicated than the process I was following earlier. Is there a quicker/simpler solution?

 

[eumyang]

Really? I think the above is so much simpler than your first attempt, radicals and all. I want to kick myself for not seeing it earlier.