Question

$sin\left(ta{n}^{-1}x\right),|x|<1$, is equal to
a) $\frac{x}{\sqrt{1-x^2}}$
b) $\frac{1}{\sqrt{1-x^2}}$
c) $\frac{1}{\sqrt{1+x^2}}$
d) $\frac{x}{\sqrt{1+x^2}}$

Answer 1

$sin\left(ta{n}^{-1}x\right)=sin\left[si{n}^{-1}\frac{x}{\sqrt{1+x^2}}\right][\because ta{n}^{-1}x=si{n}^{-1}\frac{x}{\sqrt{1+x^2}}]$
$=\frac{x}{\sqrt{1+x^2}}$.
Hence, the correct option is (d).