InterviewSolution
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InterviewSolution
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Find the sum `cos e c^(-1)sqrt(10)+cos e c^(-1)sqrt(50)+cos e c^(-1)sqrt(170)++cos e c^(-1)sqrt((n^2+1)(n^2+2n+2))` |
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Answer» Let `S = cosec^-1sqrt10+cosec^-1sqrt50+cosec^-1sqrt170+...+cosec^-1sqrt((n^2+1)(n^2+2n+2))` Here, `T_n = cosec^-1sqrt((n^2+1)(n^2+2n+2))` Let `cosec^-1sqrt((n^2+1)(n^2+2n+2)) = theta` `=>cosec theta = sqrt((n^2+1)(n^2+2n+2))` `=>cosec^2theta = (n^2+1)(n^2+2n+2)` `=>cosec^2theta = (n^2+1)(n^2+1+2n+1)` `=>1+cot^2theta = (n^2+1)(n^2+1)+2n(n^2+1)+n^2+1` `=>1+cot^2theta = (n^2+1+n)^2+1` `=>cot^2theta = (n^2+1+n)^2` `=>tan theta = 1/(1+n^2+n) = ((n+1) - n)/(1+n(n+1))` `=>theta = tan^-1(((n+1) - n)/(1+n(n+1)))` `=>theta = tan^-1(n+1) - tan^-1(n)` `:. T_n = tan^-1(n+1) - tan^-1(n)` Now, `S = tan^-1(2) - tan^-1(1) + tan^-1(3) - tan^-1(2)+tan^-1(4) - tan^-1(3)+...+tan^-1(n+1) - tan^-1(n)` `=>S = tan^-1(n+1) - tan^-1(1)` `=>S = tan^-1(n+1) - pi/4.` |
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