Simplify sin^-1(1-x^2/1+x^2), 0<x<1?

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Answer 1 · 2018-04-26T17:17:29

# pi/2-2tan^-1x, x in (0,1)#.

Let, #f(x)=sin^-1{(1-x^2)/(1+x^2)}#.

We know that, #(1-tan^2theta)/(1+tan^2theta)=cos2theta#.

So, sub.ing #x=tantheta#, we get,

#sin^-1{(1-x^2)/(1+x^2)}#,

#=sin^-1{(1-tan^2theta)/(1+tan^2theta)}#,

#=sin^-1(cos2theta)#.

#:. f(x)=sin^-1{sin(pi/2-2theta)}............(ast_1)#.

Given that #0 lt x lt 1:. 0 lt tantheta lt 1, or, #

#tan0 lt tantheta lt tan(pi/4)#.

Recall that, #tan# is #uarr" in "Q_1=(0,pi/2)#.

#:. tan0 lt tantheta lt tan(pi/4) rArr 0 lt theta lt pi/4#.

Multiplying by #2 gt 0, 0 lt 2theta lt pi/2#

# :. (pi/2-2theta) in [0,pi/2]............(ast_2)#.

Note that, #sin^-1(sinalpha)=alpha hArr alpha in [-pi/2,pi/2]...(ast_3)#.

From #(ast_1), (ast_2), and (ast_3)#, we have,

# f(x)=pi/2-2theta, i.e., #

# sin^-1{(1-x^2)/(1+x^2)}=pi/2-2tan^-1x, x in (0,1)#.

Enjoy Maths!