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Integration by Partial Fractions

Indicate the ...

Question

Indicate the relation which can hold in their respective domain for infinite values of $x$ .

A

tan ∣ ∣ {tan}^{- 1} x ∣ ∣ = | x |

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B

cot ∣ ∣ {cot}^{- 1} x ∣ ∣ = | x |

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C

{tan}^{- 1} | tan x | = | x |

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D

sin ∣ ∣ {sin}^{- 1} x ∣ ∣ = | x |

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Solution

The correct options are
A $tan ∣ ∣ {tan}^{- 1} x ∣ ∣ = | x |$
B $cot ∣ ∣ {cot}^{- 1} x ∣ ∣ = | x |$
C $sin ∣ ∣ {sin}^{- 1} x ∣ ∣ = | x |$
D ${tan}^{- 1} | tan x | = | x |$
$t a n (t a n^{- 1} (x))$
for $x < 0$ $t a n (t a n^{- 1} (x))$ implies
$= t a n (- t a n^{- 1} (x))$
$= - t a n (t a n^{- 1} (x))$
$= - x$ ...(i)
for $x > 0$
$t a n (t a n^{- 1} (x))$ implies
$= t a n (t a n^{- 1} (x))$
$= x$ ...(ii)
Hence
$t a n (| t a n^{- 1} (x) |) = | x |$
$c o t (c o t^{- 1} (x))$
for $x < 0$ $c o t (c o t^{- 1} (x))$ implies
$= c o t (π - c o t^{- 1} (x))$
$= - c o t (c o t^{- 1} (x))$
$= - x$
for $x > 0$
$c o t (c o t^{- 1} (x))$ implies
$= c o t (c o t^{- 1} (x))$
$= x$ ...(ii)
Hence
$c o t (| c o t^{- 1} (x) |) = | x |$
$s i n (s i n^{- 1} (x))$
for $x < 0$ $s i n (s i n^{- 1} (x))$ implies
$= s i n (- s i n^{- 1} (x))$
$= - s i n (s i n^{- 1} (x))$
$= - x$
for $x > 0$
$s i n (s i n^{- 1} (x))$ implies
$= s i n (s i n^{- 1} (x))$
$= x$
Hence
$s i n (| s i n^{- 1} (x) |) = | x |$
$t a n^{- 1} (t a n (x))$
For $x ϵ (\frac{π}{2}, π] \cup [\frac{- 3 π}{2}, 2 π)$
$t a n^{- 1} (t a n (- x))$
$= t a n^{- 1} (- t a n (x))$
$= - t a n^{- 1} (t a n (x))$
$= - x$
For $x ϵ [0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$
$t a n^{- 1} (t a n (x))$
$= x$
Hence $t a n^{- 1} | t a n (x) | = | x |$

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Similar questions

Q. Number of false relations of the following

(i) t a n | t a n^{- 1} x | = | x | (i i) c o t | c o t^{- 1} x | = | x | (i i i) t a n^{- 1} | t a n x | = | x | (i v) s i n | s i n^{- 1} x | = | x |

Q. The set of values of

x

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

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

{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 value of x for which

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

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