The set of values of x for which tan−1⁡x1−x2=sin−1⁡x holds, is

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
$[- 1, 0]$

Solution:

We observe that RHS is defined for all $x \in [- 1, 1]$

whereas LHS is meaningful for $- 1 < x < 1$

Also, for $x \in (- 1, 1)$ , we have

$\begin{aligned} \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) \end{aligned}$

$= \tan^{- 1} (\frac{\sin θ}{\cos θ})$ , where $x = \sin θ$

$= θ = \sin^{- 1} x$

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