Express tan-1x-tan-12x-1-x2 in the terms of tan-13x-x3-1-3x2

#160;tan ^ 1 x +tan ^ 1 2x / 1 x ^ 2 #160; =tan ^ 1 ( x+ 2x / 1 x ^ 2 #160;/ 1 x ( 2x / 1 x ^ 2 #160;) #160; #160;) #160; ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

Express tan...

Question

Express $t a n^{- 1} x + t a n^{- 1} \frac{2 x}{1 - x^{2}}$ in the terms of $t a n^{- 1} \frac{3 x - x^{3}}{1 - 3 x^{2}}$

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Solution

${tan}^{- 1} x + {tan}^{- 1} \frac{2 x}{1 - x^{2}} = {tan}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{x + \frac{2 x}{1 - x^{2}}}{1 - x (\frac{2 x}{1 - x^{2}})} ⎞ ⎟ ⎟ ⎠$ using $t a n^{- 1} x + t a n^{- 1} y = t a n^{- 1} \frac{x + y}{1 - x y}$
$= {tan}^{- 1} (\frac{x - x^{3} + 2 x}{1 - x^{2} - 2 x^{2}}) = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$

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Similar questions

Q. If

y = {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) + {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) - {tan}^{1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}})

then Show that

\frac{d y}{d x} = \frac{1}{1 + x^{2}}

Q. If

3 \cdot {tan}^{- 1} x = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})

, then

x

belongs to

Q. If

x > \frac{1}{\sqrt{3}}

then 3

t a n^{- 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. 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}}

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Express tan−1x+tan−12x1−x2 in the terms of tan−13x−x31−3x2

tan−1x+tan−12x1−x2=tan−1⎛⎜ ⎜⎝x+2x1−x21−x(2x1−x2)⎞⎟ ⎟⎠ using tan−1x+tan−1y=tan−1x+y1−xy=tan−1(x−x3+2x1−x2−2x2)=tan−1(3x−x31−3x2)

Express $t a n^{- 1} x + t a n^{- 1} \frac{2 x}{1 - x^{2}}$ in the terms of $t a n^{- 1} \frac{3 x - x^{3}}{1 - 3 x^{2}}$