How to show #sin^(-1)(x)=tan^(-1)((x)/(1-x^2)^(1/2))# ?

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Answer 1 · 2018-06-03T04:45:28

Let #x=sintheta#

So #RHS=tan^-1(x/(1-x^2)^(1/2))#

#=tan^-1(sintheta/(1-sin^2theta)^(1/2))#

#=tan^-1(sintheta/(cos^2theta)^(1/2))#

#=tan^-1(sintheta/costheta)#

#=tan^-1tantheta#

#=theta=sin^-1x=LHS#