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Find the sum cosec^{-1}{10}+cosec^{-1}{50}+cosec^{-1}{170}+...

Question

Find the sum $cose c^{- 1} 10 + cose c^{- 1} 50 + cose c^{- 1} 170 + ... + cose c^{- 1} (n^{2} + 1) (n^{2} + 2 n + 2) .$

tan^{- 1} (n + 1) - \frac{π}{4}

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Find the sum

cose c^{- 1} 10 + cose c^{- 1} 50 + cose c^{- 1} 170 + ... + cose c^{- 1} (n^{2} + 1) (n^{2} + 2 n + 2) .

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Text solutionVerified

Let

θ = cose c^{- 1} (n^{2} + 1) (n^{2} + 2 n + 2)

\Rightarrow cose c^{2} θ = (n^{2} + 1) (n^{2} + 2 n + 2) = (n^{2} + 1)^{2} + 2 n (n^{2} + 1) + n^{2} + 1 = (n^{2} + n + 1)^{2} + 1

\Rightarrow co t^{2} θ = (n^{2} + n + 1)^{2}

\Rightarrow t an θ = \frac{1}{( n ^{2} + n + 1 )} = \frac{( n + 1 ) - n}{1 + ( n + 1 ) n}

\Rightarrow θ = tan^{- 1} \frac{1}{( n ^{2} + n + 1 )} = \frac{( n + 1 ) - n}{1 + ( n + 1 ) n} = tan^{- 1} (n + 1) - t a n^{- 1} n

Thus, the sum of n terms of the given series

= (t a n^{- 1} 2 - t a n^{- 1} 1) + (t a n^{- 1} 3 - t a n^{- 1} 2) + (t a n^{- 1} 4 - t a n^{- 1} 3) + ... + tan^{- 1} (n + 1) - t a n^{- 1} n

= tan^{- 1} (n + 1) - \frac{π}{4}

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Question Text	Find the sum $cose c^{- 1} 10 + cose c^{- 1} 50 + cose c^{- 1} 170 + ... + cose c^{- 1} (n^{2} + 1) (n^{2} + 2 n + 2) .$
Answer Type	Text solution:1
Upvotes	150