Find the value of x- if tan -11- x 1- x -12 tan -1 x - x -0A- 1B- 13C- 3D- No solution

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

Find the valu...

Question

Find the value of x, if
${tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, x > 0 .$

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\frac{1}{\sqrt{3}}

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\sqrt{3}

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No solution

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Solution

The correct option is B $\frac{1}{\sqrt{3}}$
${tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x \Rightarrow 2 {tan}^{- 1} (\frac{1 - x}{1 + x}) = {tan}^{- 1} x We know that {tan}^{- 1} x - {tan}^{- 1} y = {tan}^{- 1} (\frac{x - y}{1 + x . y}) So, we can write {tan}^{- 1} (\frac{1 - x}{1 + x}) = {tan}^{- 1} 1 - {tan}^{- 1} x Therefore, 2 [{tan}^{- 1} 1 - {tan}^{- 1} x] = {tan}^{- 1} x \Rightarrow 3 {tan}^{- 1} x = 2 (\frac{π}{4}) \Rightarrow {tan}^{- 1} x = \frac{π}{6} \Rightarrow x = tan \frac{π}{6} \Rightarrow x = \frac{1}{\sqrt{3}}$

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Q. Solve the following equations for x:
(i)

(ii)
(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}

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(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

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(vii)

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(ix)

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(i) tan⁻¹2x + tan⁻¹3x = nπ +

\frac{3 π}{4}

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹

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(iv) tan⁻¹

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

tan⁻¹x = 0, where x > 0
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(viii)

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

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(x)

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

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

Find the value of x, If

{tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, x > 0

Q. Find the value of x, if

{tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, x > 0 .

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Find the value of x, if tan−1 (1−x1+x) = 12tan−1 x, x> 0.

Find the value of x, if
${tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, x > 0 .$