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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Solve tan-1...

Question

Solve ${tan}^{- 1} x + {tan}^{- 1} 2 x = n π - {tan}^{- 1} 3 x, n ϵ I, n ϵ R$ ,for $n$ and $x$

A

x = 0, n = 0

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B

x = 1, n = 1

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C

x = - 1, n = - 1

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D

x = 2, n = 2

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Solution

The correct options are
A $x = 0, n = 0$
B $x = 1, n = 1$
C $x = - 1, n = - 1$
${tan}^{- 1} (x) + {tan}^{- 1} (2 x)$
$= {tan}^{- 1} (\frac{3 x}{1 - 2 x^{2}})$
$= n π - {tan}^{- 1} 3 x$
Hence, $\frac{3 x}{1 - 2 x^{2}} = tan (n π - t a n^{- 1} (3 x))$
Now for all $n ϵ I^{+}$ we get
$\frac{3 x}{1 - 2 x^{2}} = tan (n π - {tan}^{- 1} (3 x))$
$= - tan ({tan}^{- 1} (3 x))$
$= - 3 x$
Hence
$\frac{3 x}{1 - 2 x^{2}} = - 3 x$
$3 x [\frac{1}{1 - 2 x^{2}} + 1] = 0$
$3 x [\frac{2 - 2 x^{2}}{1 - 2 x^{2}}] = 0$
$6 x [\frac{1 - x^{2}}{1 - 2 x^{2}}] = 0$
Hence $x = 0$ and $x^{2} - 1 = 0$
$x = \pm 1$ .
Hence for all $n ϵ {0, 1, 2, . . . N}$ we get $x = - 1, 0, 1$ .
Now for $n ϵ I^{-}$ we get
$\frac{3 x}{1 - 2 x^{2}} = - {tan}^{- 1} (n π + {tan}^{- 1} (3 x))$
$3 x [\frac{2 - 2 x^{2}}{1 - 2 x^{2}}] = 0$
$6 x [\frac{1 - x^{2}}{1 - 2 x^{2}}] = 0$
Hence for $n ϵ {- 1, - 2, - 3... - N}$ $x = {- 1, 0, 1}$ .

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Similar questions

Q. 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(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

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

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

\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}

l i m n \to \infty \sum_{r = 1}^{n} \frac{n}{n^{2} + r^{2} x^{2}}, x > 0

{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{n}{4}

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

Prove that $\frac{1}{2} t a n (\frac{x}{2}) + \frac{1}{4} t a n (\frac{x}{4}) + . . . + \frac{1}{2^{n}} t a n (\frac{x}{2^{n}}) = \frac{1}{2^{n}} c o t + (\frac{x}{2^{n}}) - c o t x$ for all $n ϵ N$ and $0 < x < \frac{x}{2}$ .

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