Prove the following:
$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}$

Solution

$L . H . S . = \tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{8}) = \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{5}}{1 - (\frac{1}{3}) (\frac{1}{5})}) + \tan^{- 1} (\frac{\frac{1}{7} + \frac{1}{8}}{1 - (\frac{1}{7}) (\frac{1}{8})}) = \tan^{- 1} [\frac{\frac{5 + 3}{15}}{\frac{15 - 1}{15}}] + \tan^{- 1} [\frac{\frac{8 + 7}{56}}{\frac{56 - 1}{56}}] = \tan^{- 1} [\frac{\frac{8}{15}}{\frac{14}{15}}] + \tan^{- 1} [\frac{\frac{15}{56}}{\frac{55}{56}}] = \tan^{- 1} (\frac{8}{14}) + \tan^{- 1} (\frac{15}{55}) = \tan^{- 1} (\frac{4}{7}) + \tan^{- 1} (\frac{3}{11}) = \tan^{- 1} (\frac{\frac{4}{7} + \frac{3}{11}}{1 - (\frac{4}{7}) (\frac{3}{11})}) = \tan^{- 1} (\frac{\frac{44 + 21}{77}}{\frac{77 - 12}{77}}) = \tan^{- 1} (\frac{65}{65}) = \tan^{- 1} 1 (\tan \frac{π}{4}) = \frac{π}{4}$

= R.H.S.

Hence proved.

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Inverse Trigonometric Functions

Prove the following:
$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}$

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Prove the following:
$\tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{8}$