Prove that: tan−1(21)+tan−1(51)+tan−1(81)=4π

=tan−1(21)+tan−1(51)+tan−1(81)=tan−11−21×5121+51+tan−1(81) [∵21×51=101<1]=tan−1(97)+tan−1(81)=tan−1⎝⎛1−97×8197+81⎠⎞=tan−1(7265×6572)=tan−1(1)=4π=RHS

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Prove that: ^{-1} ({1}{2}) + ^{-1} ({1}{5}) + ^{-1} ({1}{8}) =

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Prove that: $tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}$

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Prove that:

tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}

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13 mins ago

= tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8})

= tan^{- 1} \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \times \frac{1}{5}} + tan^{- 1} (\frac{1}{8})

[∵ \frac{1}{2} \times \frac{1}{5} = \frac{1}{10} < 1]

= tan^{- 1} (\frac{7}{9}) + tan^{- 1} (\frac{1}{8}) = tan^{- 1} ⎝ ⎛ \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \times \frac{1}{8}} ⎠ ⎞ = tan^{- 1} (\frac{65}{72} \times \frac{72}{65})

= tan^{- 1} (1) = \frac{π}{4} = R H S

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Question Text	Prove that: $tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}$
Answer Type	Text solution:1
Upvotes	150