The value of x for which
${tan}^{- 1} x + {sin}^{- 1} x = {tan}^{- 1} 2 x$ is

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Solution

${tan}^{- 1} x + {sin}^{- 1} x = {tan}^{- 1} 2 x \Rightarrow {sin}^{- 1} x = {tan}^{- 1} 2 x - {tan}^{- 1} x \Rightarrow {tan}^{- 1} \frac{x}{\sqrt{1 - x^{2}}} = {tan}^{- 1} (\frac{2 x - x}{1 + 2 x^{2}}] \Rightarrow \frac{x}{\sqrt{1 - x^{2}}} = \frac{x}{1 + 2 x^{2}} \Rightarrow 2 x^{3} + x = x \sqrt{1 - x^{2}} \Rightarrow 2 x^{3} - x \sqrt{1 - x^{2}} + x = 0 \Rightarrow x (2 x^{2} - \sqrt{1 - x^{2}} + 1) = 0 \Rightarrow x = 0 o r 2 x^{2} + 1 = \sqrt{1 - x^{2}} \Rightarrow x = 0 o r 4 x^{4} + 4 x^{2} + 1 = 1 - x^{2} \Rightarrow x = 0 o r 4 x^{4} + 5 x^{2} = 0 \Rightarrow x = 0 is the only solution .$

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