Find csc − 1 ( 10 ) + csc − 1 ( 50 ) + csc − 1 ( 170 ) + ⋯ , with the nth term csc − 1 ( ( n 2 + 1 ) ( n 2 + 2 n + 2 ) ) .

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MathematicsInverse Trigonometric FunctionsSummation of seriesHard2 minQB

MathematicsHardsingle choice

Find $csc^{- 1} (10) + csc^{- 1} (50) + csc^{- 1} (170) + \dots$ , with the nth term $csc^{- 1} ((n^{2} + 1) (n^{2} + 2 n + 2))$ .

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Solution:

Since (n²+1)(n²+2n+2)=1+(n²+n+1)², the nth $csc^{- 1} t er m$ is $cot^{- 1} (n^{2} + n + 1) = tan^{- 1} (n + 1) - tan^{- 1} n$ . The partial sum $tan^{- 1} (n + 1) - tan^{- 1} 1$ tends to $π /4$ .

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

⏱ 2mℹ️ Source: QB

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