If x - 1 or x -1- then cosec-1x-A- None of theseB- -1-2-xC- sin - 11- x D- 1-sin -1x

The correct option is C 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) Or sin(y) = 1/x y = sin^ 1 ( 1 ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If x ⩾ 1 or x...

Question

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

s e c^{- 1} (\frac{π}{2} - x)

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

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

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None of these

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Solution

The correct option is B $s i n^{-} 1 (\frac{1}{x})$
$L e t c o s e c^{- 1} (x) = y S i n c e, x ⩾ 1 o r x ⩽ - 1; w e c a n w r i t e - x = c o s e c (y) x = \frac{1}{s i n (y)} O r s i n (y) = \frac{1}{x} y = s i n^{- 1} (\frac{1}{x})$

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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) =

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

Q. The inverse of the function is

f (x) = \frac{10^{x} - 10^{- x}}{10^{x} + 10^{- x}} + 1

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)

Q. If

f : R \to (- 1, 1)

is defined by

f (x) = \frac{- x | x |}{1 + x^{2}}, then f^{- 1} (x)

equals
(a)

\sqrt{\frac{|x|}{1 - |x|}}

(b)

S g n (x) \sqrt{\frac{|x|}{1 - |x|}}

(c)

- \sqrt{\frac{x}{1 - x}}

(d) None of these

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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⩾1orx⩽−1; we can write−x=cosec(y)x=1sin(y)Or sin(y)=1xy=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 $s i n^{-} 1 (\frac{1}{x})$
$L e t c o s e c^{- 1} (x) = y S i n c e, x ⩾ 1 o r x ⩽ - 1; w e c a n w r i t e - x = c o s e c (y) x = \frac{1}{s i n (y)} O r s i n (y) = \frac{1}{x} y = s i n^{- 1} (\frac{1}{x})$