If x - 1 or x -1- then cosec-1x-A- -1-2-xB- sin -1 I - x C- 1-sin -D- cos -11-x

The correct option is B sin^ 1 ( 1/x )Let cosec^ 1(x) = y Since, x ⩾ 1 or x ⩽ 1 we can write x = cosec(y) ⇒ x = 1/sin(y) ⇒sin y = 1/x ⇒ y = sin^ 1 ( 1/x ) ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If x ⩾ 1 or x...

Question

$I f x ⩾ 1 o r x ⩽ - 1, t h e n c o s e c^{- 1} (x) =$

{sec}^{- 1} (\frac{π}{2} - x)

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

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

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

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Solution

The correct option is B ${sin}^{- 1} (\frac{1}{x})$
$L e t c o s e c^{- 1} (x) = y S i n c e, x ⩾ 1 o r x ⩽ - 1$
we can write
$x = c o s e c (y) \Rightarrow x = \frac{1}{sin (y)} \Rightarrow sin y = \frac{1}{x} \Rightarrow y = {sin}^{- 1} (\frac{1}{x})$

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Similar questions

I f x ⩾ 1 o r x ⩽ - 1, t h e n c o s e c^{- 1} (x) =

\frac{d}{d x} [\sqrt{cos \sqrt{x}}] =

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

{s i n}^{- 1} (1 - x) - 2 {s i n}^{- 1} x = π / 2

.
(b) If

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

, then prove that x is equal to

0, 1 / 2

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

Q. Assertion :

{cosec}^{- 1} (3 / 2) + {cos}^{- 1} (2 / 3) - 2 {cot}^{- 1} (1 / 7) - {cot}^{- 1} 7

is equal to

{cot}^{- 1} 7

Reason:

{sin}^{- 1} x + {cos}^{- 1} x = π / 2, {tan}^{- 1} x + {cot}^{- 1} x = π / 2, {cosec}^{- 1} x = {sin}^{- 1} (1 / x),

{cot}^{- 1} (1 / x) = {tan}^{- 1} (x), {cot}^{- 1} (x) = {tan}^{- 1} (1 / x)

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If x⩾1 or x ⩽−1, then cosec−1(x)=

The correct option is B sin−1(1x)Let cosec−1(x)=ySince,x⩾1 or x⩽−1 we can write x=cosec(y)⇒x=1sin(y)⇒siny=1x⇒y=sin−1(1x)

$I f x ⩾ 1 o r x ⩽ - 1, t h e n c o s e c^{- 1} (x) =$

The correct option is B ${sin}^{- 1} (\frac{1}{x})$
$L e t c o s e c^{- 1} (x) = y S i n c e, x ⩾ 1 o r x ⩽ - 1$
we can write
$x = c o s e c (y) \Rightarrow x = \frac{1}{sin (y)} \Rightarrow sin y = \frac{1}{x} \Rightarrow y = {sin}^{- 1} (\frac{1}{x})$