First slide

Properties of ITF

Question

If $- 1 < x < 0,$ then $\sin^{- 1} x$ equals

Moderate

A

$π - \sin^{- 1} \sqrt{1 - x^{2}}$
B

$\tan^{- 1} \frac{x}{\sqrt{1 - x^{2}}}$
C

$- \cot^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})$
D

none of these

Solution

Let $\sin^{- 1} x = θ . Then, x = \sin θ$
Now, $- 1 < x < 0 \Rightarrow - \frac{π}{2} < θ < 0$
$\begin{array}{l} ∴ π - \sin^{- 1} \sqrt{1 - x^{2}} \\ = π - \sin^{- 1} (\cos θ) \\ = π - \sin (\frac{π}{2} + θ) = π - (\frac{π}{2} + θ) = \frac{π}{2} - θ \\ As - \frac{π}{2} < θ < 0 \\ ∴ 0 < π - θ < \frac{π}{2} \Rightarrow 0 < π - \sin^{- 1} \sqrt{1 - x^{2}} < \frac{π}{2} \\ ∴ \sin^{- 1} x \neq π - \sin^{- 1} \sqrt{1 - x^{2}} \end{array}$
So, option (a) is not correct.
We have,
$- \cot^{- 1} (\frac{\sqrt{1 - x^{2}}}{x}) = - \cot^{- 1} (\frac{\cos θ}{\sin θ}) = - \cot^{- 1} (\cot θ) \begin{array}{l} = \cot^{- 1} (\cot (- θ)) \\ = - θ \\ = - \sin^{- 1} x \end{array} [∵ θ \in (- π / 2, 0) \Rightarrow - θ \in (0, π / 2)]$
Thus, option (c) is also not correct.

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