The set of values of x for which $tan^{-1}\frac{x}{\sqrt{1-x^2}}= sin^{-1} x $ holds , is
Answer & explanation
Correct answer: option 2
We observe that RHS is defined for all x ∈ [-1,1]
whereas LHS is meaningful for -1 < x < 1.
Also, for x ∈(-1, 1), we have
$tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
$=tan^{-1}\left(\frac{sin \theta }{\cos\theta }\right)$, where x = sin θ
$= θ = sin^{-1} x $