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Complex Numbers

If x takes ...

Question

If $x$ takes negative permissible value, then $s i n^{- 1} (- x)$ is equal to

A

c o s^{- 1} \sqrt{1 - x^{2}}

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B

- c o s^{- 1} \sqrt{1 - x^{2}}

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C

c o s^{- 1} \sqrt{x^{2} - 1}

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D

π - c o s^{- 1} \sqrt{1 - x^{2}}

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Solution

The correct option is B $- c o s^{- 1} \sqrt{1 - x^{2}}$
$x$ takes negative permissible value, then
${sin}^{- 1} (- x)$
$= - {sin}^{- 1} (x)$
$= - {cos}^{- 1} (\sqrt{1 - x^{2}})$

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Similar questions

{cos}^{- 1} (x^{2} + \frac{1}{x^{2}} - 1) + {sin}^{- 1} (x^{2} - \frac{1}{x^{2}}) + {tan}^{- 1} (x^{2})

(where x \in R - {0})

- 1 \leq x \leq 0

sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} - {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}

0 \leq x \leq 1

sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}

Q. Assertion :Derivative of

{sin}^{- 1} (\frac{2 x}{1 + x^{2}})

{cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

1,

0 < x < 1

{sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

- 1 \leq x \leq 1

Q. Assertion :Derivative of

s i n^{- 1} (\frac{2 x}{1 + x^{2}})

with respect to

c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

1

0 < x < 1.

s i n^{- 1} (\frac{2 x}{1 + x^{2}}) = c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

- 1 \leq x \leq 1

.

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