First slide

Properties of ITF

Question

If $x$ takes negative permissible value, then $\sin^{- 1} x =$

Moderate

A

$\cos^{- 1} \sqrt{1 - x^{2}}$
B

$- \cos^{- 1} \sqrt{1 - x^{2}}$
C

$\cos^{- 1} \sqrt{x^{2} - 1}$
D

$π - \cos^{- 1} \sqrt{1 - x^{2}}$

Solution

Let $\sin^{- 1} x = y .$ Then, $x = \sin y$
Now,
$- 1 \leq x \leq 0 \Rightarrow - \frac{π}{2} \leq \sin^{- 1} x \leq 0 \Rightarrow - \frac{π}{2} \leq y \leq 0 .$
We have,
$\cos y = \sqrt{1 - \sin^{2} y}$
$\Rightarrow \cos y = \sqrt{1 - x^{2}}$ for $0 \leq y \leq π .$ …(i)
Now,
$- \frac{π}{2} \leq y \leq 0$
$\Rightarrow \frac{π}{2} \geq - y \geq 0$
$\Rightarrow \cos (- y) = \sqrt{1 - x^{2}}$ [From (i)]
$\Rightarrow - y = \cos^{- 1} \sqrt{1 - x^{2}} \Rightarrow y = - \cos^{- 1} \sqrt{1 - x^{2}}$

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