Solve tan-1 x - -1 x-

Given, tan ^ 1 x gt; ^ 1 x ⇒tan( tan ^ 1 x #160;) #160; gt;tan( ^ 1 x #160;) #160; #160;⇒tan( tan ^ 1 x #160;) #160; ...

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Integration Using Substitution

Solve tan-1...

Question

Solve ${tan}^{- 1} x > {cot}^{- 1} x$ .

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Solution

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

\frac{π}{12}

(\frac{8}{79})

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

\frac{3 π}{4}

\frac{8}{31}

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

\frac{π}{12}

(\frac{8}{79})

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

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

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

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

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

{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}}

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

x

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

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