If 4 cos ⁡ 36 o + c o t 7 1 2 ∘ = n 1 + n 2 + n 3 + n 4 + n 5 + n 6 , then the value of ∑ i = 1 6 n i 2 10 is equal to

Question

If 4 cos ⁡ 36 o + c o t 7 1 2 ∘ = n 1 + n 2 + n 3 + n 4 + n 5 + n 6 , then the value of ∑ i = 1 6 n i 2 10 is equal to,

MARKS App · Accepted Answer

4 cos ⁡ 36 o = 4 5 + 1 4 = 5 + 1 cot ⁡ 7 1 2 o = 1 + cos ⁡ 15 ° sin ⁡ 15 ° = 1 + 3 + 1 2 2 3 - 1 2 2 = 2 2 + 3 + 1 3 - 1 = 2 2 + 3 + 1 3 + 1 3 - 1 3 + 1 = 2 6 + 2 2 + 3 + 3 + 3 + 1 2 = 2 6 + 2 3 + 2 2 + 4 2 = 6 + 3 + 2 + 2 = 6 + 4 + 3 + 2 ⇒ 4 cos ⁡ 36 o + c o t 7 1 2 o = 5 + 1 + 6 + 4 + 3 + 2 = 1 + 2 + 3 + 4 + 5 + 6 = n 1 + n 2 + n 3 + n 4 + n 5 + n 6 ⇒ ∑ i = 1 6 n i 2 = 1 2 + 2 2 + 3 3 + 4 2 + 5 2 + 6 2 = 6 × 7 × 13 6 = 9 1 ⇒ ∑ i = 1 6 n i 2 10 = 91 10 = 9 . 1

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