Solve- 2tan - 134 - tan - 11731

2tan^ 1(34) tan^ 1(711) tan^ 1(34)+{tan^ 1(34) tan^ 1(711)} =tan^ 1(34)+tan^ 1(34 7111+34.711) =tan^ 1(34)+tan^ 1(33 284444+2144) =tan^ ...

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Domain and Range of Basic Inverse Trigonometric Functions

Solve: 2tan ...

Question

Solve:
$2 {tan}^{- 1} \frac{3}{4} - {tan}^{- 1} \frac{17}{31}$

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2 {sin}^{- 1} \frac{3}{5} - {tan}^{- 1} \frac{17}{31} = \frac{π}{4}

2 {sin}^{- 1} (\frac{3}{5}) - {tan}^{- 1} (\frac{17}{31}) = \frac{π}{4}

Solve the following for $x : s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}$

Show that: $2 s i n^{- 1} (\frac{3}{5}) - t a n^{- 1} (\frac{17}{31}) = \frac{π}{4}$

\tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{13} = \tan^{- 1} \frac{2}{9}

\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \frac{1}{2} \cos^{- 1} \frac{3}{5} = \frac{1}{2} \sin^{- 1} (\frac{4}{5})

\tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \frac{12}{5}

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\sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16} = π

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2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{4}{7}

2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})

2 \sin^{- 1} \frac{3}{5} = \tan^{- 1} \frac{24}{7}

\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \frac{1}{2} \cos^{- 1} \frac{3}{5} = \frac{1}{2} \sin^{- 1} (\frac{4}{5})

\tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \frac{12}{5}

\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{4}

\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{2}

2 \sin^{- 1} \frac{3}{5} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{4}{7}

2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})

4 \tan^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = \frac{π}{4}

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