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

Jun 26, 2017

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

#### Explanation:

Given: $\tan \left({\sin}^{-} 1 \left(x\right)\right) = \frac{x}{\sqrt{1 - {x}^{2}}}$

Change the $\tan$ into $\frac{\sin}{\cos}$

$\sin \frac{{\sin}^{-} 1 \left(x\right)}{\cos} \left({\sin}^{-} 1 \left(x\right)\right) = \frac{x}{\sqrt{1 - {x}^{2}}}$

The numerator becomes x by definition:

$\frac{x}{\cos} \left({\sin}^{-} 1 \left(x\right)\right) = \frac{x}{\sqrt{1 - {x}^{2}}}$

Substitute $\sqrt{1 - {u}^{2}}$ for $\cos \left(u\right)$

$\frac{x}{\sqrt{1 - {\sin}^{2} \left(\left({\sin}^{-} 1 \left(x\right)\right)\right)}} = \frac{x}{\sqrt{1 - {x}^{2}}}$

${\sin}^{2} \left({\sin}^{-} 1 \left(x\right)\right)$ becomes ${x}^{2}$:

$\frac{x}{\sqrt{1 - {x}^{2}}} = \frac{x}{\sqrt{1 - {x}^{2}}}$ Q.E.D.

Jun 26, 2017

Let ${\sin}^{-} 1 x = \theta$

so $\sin \theta = x$
and
$\tan \theta = \sin \frac{\theta}{\cos} \theta = \sin \frac{\theta}{\sqrt{1 - {\sin}^{2} \theta}} = \frac{x}{\sqrt{1 - {x}^{2}}}$

$\implies \tan \left({\sin}^{-} 1 x\right) = \frac{x}{\sqrt{1 - {x}^{2}}}$