Question

The set of values of $x$ for which
${\tan }^{-1}\frac{x}{\sqrt{1-x^2}}={\sin }^{-1}x$ holds, is

• A. $R$
• B. $(-1,1)$
• C. $[0,1]$
• D. $[-1,0]$

Answer 1

Answer: (B)

$(-1,1)$

${\sin }^{-1}x$ is defined for $x\in [-1,1]$

But for RHS $\sqrt{1-x^2}>0\Rightarrow x\in (-1,1)$

Put $x=\sin \theta \Rightarrow RHS={\tan }^{-1}\frac{sin\theta }{cos\theta }=\theta LHS={\sin }^{-1}\sin \theta =\theta LHS=RHS\text{(For every value of x in set )}$

$\therefore$ set of the values of $x\in (-1,1)$