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Question Number 146300 by akolade last updated on 12/Jul/21

Answered by nimnim last updated on 12/Jul/21

cosB=((a^2 +c^2 −b^2 )/(2ac)) =(((n^2 −1)^2 +(n^2 −2n)^2 −(n^2 −n+1)^2 )/(2(n^2 −1)(n^2 −2n))) =(((n^2 −1)^2 +(n^2 −2n+n^2 −n+1)(n^2 −2n−n^2 +n−1))/(2(n^2 −1)(n^2 −2n))) =(((n^2 −1)^2 +(2n^2 −3n+1)(−n−1))/(2(n^2 −1)(n^2 −2n))) =(((n^2 −1)^2 −(2n−1)(n−1)(n+1))/(2(n^2 −1)(n^2 −2n))) =(((n^2 −1){(n^2 −1)−(2n−1)})/(2(n^2 −1)(n^2 −2n))) =(((n^2 −1−2n+1))/(2(n^2 −2n)))=(((n^2 −2n))/(2(n^2 −2n))) cosB=(1/2) ★

cosB=a2+c2−b22ac=(n2−1)2+(n2−2n)2−(n2−n+1)22(n2−1)(n2−2n)=(n2−1)2+(n2−2n+n2−n+1)(n2−2n−n2+n−1)2(n2−1)(n2−2n)=(n2−1)2+(2n2−3n+1)(−n−1)2(n2−1)(n2−2n)=(n2−1)2−(2n−1)(n−1)(n+1)2(n2−1)(n2−2n)Missing \left or extra \right=(n2−1−2n+1)2(n2−2n)=(n2−2n)2(n2−2n)cosB=12★

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