Solve for x - tan -1 x -1-tan -1 x -tan -1 x -1-tan -1 3 x-ORProve that - Neginpmatrix frac 6 x-8 x- 31-12 x- 2lendpmatrix -tan -1 beginpmatrix 4 x 1-4 x - 2lendpmatrix -tan -1 2 x -2 x -frac1 lsqrt -3 -1

Given that tan^ 1(x 1)+tan^ 1x+tan^ 1(x+1)=tan^ 13x ⇒ tan^ 1(x 1) + tan^ 1(x+1)=tan^ 13x tan^ 1x ⇒ tan^ 1(x 1)+(x+1)/1 (x 1)(x+1)=tan^ 13x x/1+3x.x ⇒ tan^ 1 ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Solve for x :...

Question

Solve for $x : t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x .$

OR

Prove that : $(\begin{matrix} \frac{6 x - 8 x^{3}}{1 - 12 x^{2}} \end{matrix}) - t a n^{- 1} (\begin{matrix} \frac{4 x}{1 - 4 x^{2}} \end{matrix}) = t a n^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}} .$

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Solution

Given that $t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x$

$\Rightarrow t a n^{- 1} (x - 1) + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x - t a n^{- 1} x \Rightarrow t a n^{- 1} \frac{(x - 1) + (x + 1)}{1 - (x - 1) (x + 1)} = t a n^{- 1} \frac{3 x - x}{1 + 3 x . x} \Rightarrow t a n^{- 1} \frac{2 x}{2 - x^{2}} = t a n^{- 1} \frac{2 x}{1 + 3 x^{2}} \Rightarrow \frac{2 x}{2 - x^{2}} = \frac{2 x}{1 + 3 x^{2}} \Rightarrow 2 x (1 + 3 x^{2} - 2 + x^{2}) = 0 E i t h e r 2 x = 0 o r (4 x^{2} - 1) = 0 ∴ x = 0 o r x = \pm \frac{1}{2} . O R L H S = L e t y = t a n^{- 1} (\begin{matrix} \frac{6 x - 8 x^{3}}{1 - 12 x^{2}} \end{matrix}) - t a n^{- 1} (\begin{matrix} \frac{4 x}{1 - 4 x^{2}} \end{matrix}) P u t 2 x = t a n θ \Rightarrow θ = t a n^{- 1} 2 x . . . (i) ∴ y = t a n^{- 1} (\begin{matrix} \frac{3 t a n θ - t a n^{3} θ}{1 - 3 t a n^{2} θ} \end{matrix}) - t a n^{- 1} (\begin{matrix} \frac{2 t a n θ}{1 - t a n^{2} θ} \end{matrix}) \Rightarrow y = t a n^{- 1} (t a n 3 θ) - t a n^{- 1} (t a n 2 θ) \Rightarrow y = 3 θ - 2 θ = θ = t a n^{- 1} 2 x = R H S . B y (i)$

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Similar questions

Q. i) Solve for

x : {tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x

ii) Prove that

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

Solve for x the following :

(a)

{t a n}^{- 1} \frac{x - 1}{x - 2} + {t a n}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

(b)

{t a n}^{- 1} \frac{x - 1}{x + 1} + {t a n}^{- 1} \frac{2 x - 1}{2 x + 1} = {t a n}^{- 1} \frac{23}{36}

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})

, x > 0
(vii)

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

(viii)

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

(ix)

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

(x)

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

Q. Solve for X:-

{tan}^{- 1} (x + 1) + {tan}^{- 1} (x - 1) = {tan}^{- 1} 8 / 31

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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Solve for x:tan−1(x−1)+tan−1x+tan−1(x+1)=tan−13x. OR Prove that : (6x−8x31−12x2)−tan−1(4x1−4x2)=tan−12x;|2x|<1√3.

Solve for $x : t a n^{- 1} (x - 1) + t a n^{- 1} x + t a n^{- 1} (x + 1) = t a n^{- 1} 3 x .$

OR

Prove that : $(\begin{matrix} \frac{6 x - 8 x^{3}}{1 - 12 x^{2}} \end{matrix}) - t a n^{- 1} (\begin{matrix} \frac{4 x}{1 - 4 x^{2}} \end{matrix}) = t a n^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}} .$