K = 1 ∑ ∞ Tan − 1 ( 2 K 2 1 ) = θ , then Tan θ =

Tan − 1 [ 1 + ( 2 K + 1 ) ( 2 K − 1 ) ( 2 K + 1 ) − ( 2 K − 1 ) ] = Tan − 1 ( 2 K + 1 ) − Tan − 1 ( 2 K − 1 ) Expanding Σ we get, Tan − 1 1

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MathematicsInverse Trigonometric FunctionSummation of seriesMedium2 min

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$K = 1 \sum \infty Tan^{- 1} (\frac{1}{2 K ^{2}}) = θ$ , then $Tan θ =$

Answer:

Solution:

$Tan^{- 1} [\frac{( 2 K + 1 ) - ( 2 K - 1 )}{1 + ( 2 K + 1 ) ( 2 K - 1 )}] = Tan^{- 1} (2 K + 1) - Tan^{- 1} (2 K - 1)$

Expanding $Σ$ we get, $Tan^{- 1} 1$

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionSubtopic:Summation of series

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