The value of x satisfying tan - 1 1 - x1 - x - 12tan - 1x x - 0 is-

The correct option is C 1√(3) tan^ 1(1 x1+x)=12θ Let x=tanθ ⇒tan^ 1(1 tanθ1+tanθ) ⇒tan^ 1(tanπ4 tanθ1+tanπ4tanθ) tan^ 1tan(π4 θ)= ...

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Byju's Answer

Standard XII

Mathematics

Particular Solution of a Differential Equation

The value of ...

Question

The value of $x$ satisfying ${tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x (x > 0)$ is:

\frac{1}{3}

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

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\sqrt{3}

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\sqrt{2} - 1

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Solution

The correct option is C $\frac{1}{\sqrt{3}}$
${tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} θ$
Let $x = tan θ$
$\Rightarrow {tan}^{- 1} (\frac{1 - tan θ}{1 + tan θ})$
$\Rightarrow {tan}^{- 1} (\frac{tan \frac{π}{4} - tan θ}{1 + tan \frac{π}{4} tan θ})$
${tan}^{- 1} tan (\frac{π}{4} - θ) = \frac{1}{2} θ$
$\frac{π}{4} - θ = \frac{θ}{2}$
$\frac{π}{2} = 3 θ$
$θ = \frac{π}{6}$
${tan}^{- 1} x = \frac{π}{6}$
$x = \frac{1}{\sqrt{3}}$

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Q. The set of values of x satisfying the equation

\frac{\tan 3 x - \tan 2 x}{1 + \tan 3 x \tan 2 x} = 1

is______________.

Q. The value of

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