Question

How is [cot(^2)(-x)-1]/[cot(^2)x] equal to [-cot(^2)x-1]/[-cot(^2)x] ?? Why is the denominator negative also?

Answer 1

First note that:
cot x = cos x / sin x
=> cot^2 x = cos^2 x / sin^2 x
=> cot^2 (-x) = cos^2 (-x) / sin^2(-x)
= cos^2 (x) / sin^2 (x)
= cot^2 x


i.e., cot^2 (-x) = cot^2 x

Hence the LHS =
[cot(^2)(-x)-1]/[cot(^2)x] = [cot(^2)(x)-1]/[cot(^2)x]

Answer 2

Simply the RHS by removing -1 from numerator and denominator:
( -cot(^2)x-1 ) / ( -cot(^2)x ) = (cot^2 x + 1) / cot^2 x

So the LHS = RHS if and only if

cot^2 x - 1 = cot^2 x + 1 which is false, because -1 does not equal +1.

For example, let x = pi/4 or 45 degrees.

Then cot^2 x = 1^2 = 1 and
cot^2 (-x) = (-1)^2 = 1

Hence LHS = (1-1)/1 = 0
and RHS = (-1 - 1)/(-1) = 2

Answer 3

So what I'm saying is that this is not an identity; the formula does not hold.

Answer 4

... the original problem isn't? And I'm sorry but what is LHS and RHS?

Answer 5

LHS = left hand side, RHS = right ...

Answer 6

ah gotcha ha

Answer 7

So what the problem isn't legit?

Answer 8

Nope; it's not an identity.

Answer 9

Because we just showed for x = 45 degrees, it doesn't hold.

LHS = 0 but RHS = 2.

Answer 10

So the answer would not be sec(^2)x.

Answer 11

flutter man my fluttering textbook is wrong all the fluttering time it's riddled with mistakes. Thank you so much I wouldn't have suspected this one so I wouldn't have bothered checking. Here's their complete answer:

Answer 12

yea, I don't know what they're smoking.

Answer 13

It's very bad to have an error of this sort in a text book.

Answer 14

It's the Precalculus Workbook for Dummies. I have to teach myself the entire course to take the exam and I was in Europe so it's the best textbook download I could find online...

Answer 15

ironic eh