Show that tan-1 1-2 - tan-1 1-3 - -4

tan^ 1 1 2 +tan^ 1 1 3 =tan^ 1 ( 1 2 + 1 3 #160; 1 1 2 × 1 3 #160; #160; ) = tan^ 1 ( 5 6 #160; 5 6 #160; ...

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Domain and Range of Basic Inverse Trigonometric Functions

Show that t...

Question

Show that
${tan}^{- 1} (\frac{1}{2}) + {tan}^{- 1} (\frac{1}{3}) = \frac{π}{4}$

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tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{7}) + tan^{- 1} (\frac{1}{3}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}

{tan}^{- 1} \frac{1}{5} + {tan}^{- 1} \frac{1}{7} + {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{8} = \frac{π}{4}

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

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

{tan}^{- 1} \frac{1}{5} + {tan}^{- 1} \frac{1}{7} + {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{8} = \frac{π}{4}

{t a n}^{- 1} (\frac{1}{3}) + {t a n}^{- 1} (\frac{1}{5}) + {t a n}^{- 1} (\frac{1}{7}) + {t a n}^{- 1} (\frac{1}{8}) = \frac{π}{4}

t a n^{- 1} \frac{1}{3} + t a n^{- 1} \frac{1}{7} + t a n^{- 1} \frac{1}{5} + t a n^{- 1} \frac{1}{8} = \frac{π}{4} .

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