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.