Show that 2tan - 135 - tan - 1158-

Now, 2tan ^ 135 #160; = tan ^ 12×351 925 =tan^ 13016 =tan^ 1158 ...

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Trigonometric Ratios of Compound Angles

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Question

Show that $2 {tan}^{- 1} \frac{3}{5} = {tan}^{- 1} \frac{15}{8} .$

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

\tan (2 \tan^{- 1} \frac{1}{5})

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}

c o s^{- 1} (\frac{15}{7}) + 2 t a n^{- 1} (\frac{1}{5}) =

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

m

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