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
$tantheta=sintheta/costheta=sintheta/sqrt(1-sin^2theta)=x/sqrt(1-x^2)$

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

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