pipantasi4

2022-07-07

Prove that
$\mathrm{tan}\left(\mathrm{arcsin}\left(x\right)\right)=\mathrm{tan}\theta =\frac{x}{\sqrt{1-{x}^{2}}}\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}0\le x<1$

Sophia Mcdowell

Expert

The very last step can be translated to
$0\le -x<1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}\left(\mathrm{arcsin}\left(x\right)\right)=-\mathrm{tan}\left(-\mathrm{arcsin}\left(x\right)\right)=-\mathrm{tan}\left(\mathrm{arcsin}\left(-x\right)\right)=-\frac{-x}{\sqrt{1-{x}^{2}}},$
equivalent to
$-1
Oddness of the functions allows the minus sign to "cross" them.

Do you have a similar question?