Number of false relations of the following
$(i) t a n | t a n^{- 1} x | = | x | (i i) c o t | c o t^{- 1} x | = | x | (i i i) t a n^{- 1} | t a n x | = | x | (i v) s i n | s i n^{- 1} x | = | x |$ is:

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Solution

The correct option is A 1
If $x < 0, | x | = - x, t a n^{1} x < 0$
So $| t a n^{- 1} x | = - t a n^{- 1} x$ .
$∴ t a n | t a n^{- 1} x | = t a n (- t a n^{- 1} x) = - t a n t a n^{- 1} x = - x$
and for $x \leq 0, t a n | t a n^{- 1} x | = t a n t a n^{- 1} x = x$ .
$∴ t a n | t a n^{- 1} x | = | x | . ∴$ (i) is correct.
Similarly, (ii) and (iv) are correct.
In (iii), Let $x = \frac{3 π}{4}, t a n^{- 1} | t a n x | = t a n^{- 1} | - 1 |$
$= t a n^{- 1} 1 = \frac{π}{4} \neq ∣ ∣ \frac{3 π}{4} ∣ ∣ = \frac{3 π}{4}$ .
Hence, there is one false relation.

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Similar questions

Q. Assertion :Statement-1:

c o t^{- 1} (x) - t a n^{- 1} (x) > 0

for all

x < 1

Reason: Statement-2: Graph of

c o t^{- 1} (x)

is always above the graph of

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for all

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{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

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.
(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

Solve the following equations for x:
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(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) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$