If-x - 1 -or-x- -1-then-cosec-1-x- - -None of thesesec-1 - -2 - x -sin-1 - 1-x -1-sin-1-x-

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 ...

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

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

is equal to
(a)

\sin^{- 1} \sqrt{x} + C

(b)

\sin^{- 1} \{\sqrt{x} - \sqrt{x (1 - x)}\} + C

(c)

\sin^{- 1} \{\sqrt{x (1 - x)}\} + C

(d)

\sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C

Q. Write each of the following in the simplest form:

(i)

\cot^{- 1} \frac{a}{\sqrt{x^{2} - a^{2}}}, |x| > a

(ii)

\tan^{- 1} \{x + \sqrt{1 + x^{2}}\}, x \in R

(iii)

\tan^{- 1} \{\sqrt{1 + x^{2}} - x\}, x \in R

(iv)

\tan^{- 1} \{\frac{\sqrt{1 + x^{2}} - 1}{x}\}, x \neq 0

(v)

\tan^{- 1} \{\frac{\sqrt{1 + x^{2}} + 1}{x}\}, x \neq 0

(vi)

\tan^{- 1} \sqrt{\frac{a - x}{a + x}}, - a < x < a

(vii)

\tan^{- 1} \{\frac{x}{a + \sqrt{a^{2} - x^{2}}}\}, - a < x < a

(viii)

\sin^{- 1} \{\frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}}\}, - 1 < x < 1

(ix)

\sin^{- 1} \{\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}\}, 0 < x < 1

(x)

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

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

Q. If

y = x \sin^{- 1} x + \sqrt{1 - x^{2}}

, prove that

\frac{d y}{d x} = \sin^{- 1} x

I f x ⩾ 1 o r x ⩽ - 1, t h e n c o s e c^{- 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⩾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 $s i n^{-} 1 (\frac{1}{x})$
$L e t c o s e c^{- 1} (x) = y S i n c e, x ⩾ 1 o r x ⩽ - 1; w e c a n w r i t e - x = c o s e c (y) x = \frac{1}{s i n (y)} O r s i n (y) = \frac{1}{x} y = s i n^{- 1} (\frac{1}{x})$