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tan1/(1+tan1)+tan2/(1+tan2)+tan3/(1+tan3) ....................... tan88/(1+tan88)+tan89/(1+tan89) = ?

tan1/(1+tan1)+tan2/(1+tan2)+tan3/(1+tan3) ....................... tan88/(1+tan88)+tan89/(1+tan89) = ?

Grade:11

1 Answers

Aditya Gupta
2081 Points
4 years ago
we need to find S= ∑tan(r)/[1+tan(r)] where r runs from 1 to 89
S= ∑1 – 1/[1+tan(r)]= ∑1 – ∑1/[1+tan(r)]= 89 – ∑1/[1+tan(r)]
now let P= ∑1/[1+tan(r)].....(1)
write P in reverse order, we get
P= ∑1/[1+tan(90 – r)]
P= ∑1/[1+cot(r)]
P= ∑tan(r)/[1+tan(r)].........(2)
add 1 and 2
2P= ∑[1+tan(r)]/[1+tan(r)]= ∑1= 89
or P= 89/2
so, S= 89 – P= 89 – 89/2
S= 89/2
kindly approve :)

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