How do you solve #tan(sin^-1x)#=#x/(sqrt(1-x^2)#?

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Answer 1 · 2017-06-26T02:04:05

It is an identity, true for all values of x. Please see the proof below:

Given: #tan(sin^-1(x)) = x/sqrt(1-x^2)#

Change the #tan# into #sin/cos#

#sin(sin^-1(x))/cos(sin^-1(x)) = x/sqrt(1-x^2)#

The numerator becomes x by definition:

#x/cos(sin^-1(x)) = x/sqrt(1-x^2)#

Substitute #sqrt(1-u^2)# for #cos(u)#

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

#sin^2(sin^-1(x))# becomes #x^2#:

#x/sqrt(1 - x^2) = x/sqrt(1-x^2)# Q.E.D.

Answer 2 · 2017-06-26T02:05:16

Let # sin^-1x= theta#

so #sintheta=x#
and
#tantheta=sintheta/costheta=sintheta/sqrt(1-sin^2theta)=x/sqrt(1-x^2)#

#=>tan(sin^-1x)=x/sqrt(1-x^2)#