How to solve this inverse function?

Question

1192 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T19:29:32

See below.

#(sqrt[1 + x^2] + sqrt[1 - x^2])/(sqrt[1 + x^2] - sqrt[1 - x^2]) = (1 + sqrt[1 - x^4])/x^2# or

#phi = tan^-1( (1 + sqrt[1 - x^4])/x^2)# or

# (1 + sqrt[1 - x^4])/x^2 = tan phi# or

#(x^2 tan phi-1)^2= 1-x^4# or

#x^4 tan^2phi-2 x^2 tan phi+1 = 1-x^4# or

#x^2 = sin(2phi)# or

#phi = 1/2 sin^-1(x^2)#

The last step is left to the reader as an exercise. Be careful because the squaring operation can introduce strange solutions.

NOTE

#2phi = sin^-1(x^2) rArr 2phi = pi/2-cos^-1(x^2) rArr phi = pi/4-1/2cos^-1(x^2)#