How to solve this inverse function?

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1 Answer
Mar 20, 2018

Answer:

See below.

Explanation:

#(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)#