The value of cot ⁡ ∑ n = 1 19 cot - 1 ⁡ 1 + ∑ p = 1 n 2 p is:

Question

The value of cot ⁡ ∑ n = 1 19 cot - 1 ⁡ 1 + ∑ p = 1 n 2 p is:, 21 19, 19 21, 23 22, 22 23

MARKS App · Accepted Answer

We have, cot ⁡ ∑ n = 1 19 cot - 1 ⁡ 1 + ∑ p = 1 n 2 p = c o t ∑ n = 1 19 t a n - 1 1 1 + n n + 1 = c o t ∑ n = 1 19 t a n - 1 n + 1 - n 1 + n n + 1 = c o t ⁡ ∑ n = 1 19 ( t a n - 1 n + 1 - t a n - 1 n ) = cot ⁡ t a n - 1 20 - t a n - 1 1 = 1 t a n ( t a n - 1 20 - t a n - 1 1 ) = 1 20 - 1 1 + ( 20 ) . 1 = 21 19

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