Question

Write ${\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$ in term of ${\sin }^{-1}$.

Answer 1

Let

$A={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$

$\tan A=\frac{\sqrt{1-x^2}}{x}$

Using Pythagoras theorem,

$Perpendicular=\sqrt{1-x^2}$

$Base=x$

So,

$Hypotenuse=\sqrt{Bas{e}^2+Perpendicula{r}^2}$

$Hypotenuse=\sqrt{x^2+(\sqrt{1-x^2}{)}^2}$

$Hypotenuse=\sqrt{x^2+1-x^2}$

$Hypotenuse=\sqrt{1}=1$

We know that

$\sin A=\frac{Perpendicular}{Hypotenuse}$

$A={\sin }^{-1}\frac{\sqrt{1-x^2}}{1}$

$A={\sin }^{-1}\sqrt{1-x^2}$

Hence, this is the answer.