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1/1-cosO + 1/cosO = 2cosecO

2007-08-10 21:50:17 · 2 answers · asked by paulinus a 1 in Science & Mathematics Mathematics

2 answers

cosec(T) = 1/sin(T) = 1/SQRT(1 - cos^2(T))

1/(1 - cosT) + 1/cosT = 2/SQRT(1 - cos^2T)
SQRT(1 - cos^2T) = 2cosT(1 - cosT)
(1 - cosT)(1 + cosT) = 2(1 - cosT)
1 + cosT = 2 and T = 0

But the last step is not right. 1 - cosT is 0 and I have divided by 0. What has been done is take 0 = 0 in the form of (1 - cos0) and then multiplied it by something - in this case (1 + cosT)/2 but it can be anything since 0 times anything is still 0 - and set it equal to 1 - cos0. A few manipulations and you end up where you started: 1/1-cosO + 1/cosO = 2cosecO


It is similar to:
a = b
ab = b^2
ab - a^2 = b^2 - a^2
a(b - a) = (b - a)(b + a)
so a = b + a and 1 = 2 since a and be are the same
The wrong step is dividing by (b - a) which is 0.

2007-08-11 02:32:54 · answer #1 · answered by Captain Mephisto 7 · 0 0

CAN'T since

1/(1-cosO) + 1/cosO =/= 2cosecO

2007-08-11 05:42:50 · answer #2 · answered by harry m 6 · 0 0

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