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

The correct option is B $(-1,1)$

Since, x is the argument of inverse function of sin ,

$x\in (-1,1)$ $[x\ne {}_{-}^{+}1$, can be oberved from tan inverse function]

Let $x=sin\left(t\right)$ for $t\in (\frac{-\pi }{2},\frac{\pi }{2})$

Then, we can rewrite the given equation as,

$ta{n}^{-1}\frac{sint}{\sqrt{1-(sint{)}^2}}=si{n}^{-1}\left(sint\right)$

$\to ta{n}^{-1}tant=si{n}^{-1}sint$

Since $t\in (\frac{-\pi }{2},\frac{\pi }{2})$, Therefore,

$ta{n}^{-1}tant=t$ and $si{n}^{-1}sint=t$

$\to t=t$ which is always true.

So, the given equation is true for every allowed value of x i.e, $(-1,1).$

Thus, B is the correct answer.