Question

1/tan = cot right? and 1/cot = tan right?

Answer 1

Yes.

Answer 2

1/(sin/cos)
=cos/sin
=cot

Sin/cos
=tan

Answer 3

@Dido525 quick calc questoin

Answer 4

as limit x approaches 0... x1/tanx
x equals fx, and tanx equals gx

Answer 5

Proof:

Sin(x) = o/h
Cos(x) = a/h

Tan(x) = sin(x)/cos(x)
Tan(x) = o/h / a/h
=o/a

Thats the definition of tan.

Answer 6

Elaborate. I dont understand the question.

Answer 7

@Dido525 can you show me your work, i get 0 when i work it out

Answer 8

Can you use l'hopitals rule?

Answer 9

yes

Answer 10

Okay. So you have x/tan(x) .

Answer 11

Take the derivatives of the top and bottom.

Answer 12

So you get 1/sec^2(x)

Answer 13

i do all the work, just keep getting 0 as the answer!! :(

Answer 14

Which is the same as cos^2(x) .

Answer 15

Cos^2(0) =1

Answer 16

The limit is 1.

Answer 17

You probably thought cos(0) is 0. :p .

Answer 18

i end up getting sin(0)

Answer 19

could you send me the work? still stuck

Answer 20

?

1/sec^(x) is cos^2(x) not sin^2(x) .

Answer 21

how do you get 1/sec?

Answer 22

I cant. I am on my phone so it's hard to format.

Answer 23

1/tanx = cotx and derivative is -csc^2x

Answer 24

That shoukf say sec^2(x) sorry.

Answer 25

Ahh.... I see.

Answer 26

1/-csc^2x

Answer 27

But then you have x*-csc^(x) .

Answer 28

Product rule.

Answer 29

that would get cos?

Answer 30

Derivative of csc(x) is -csc(x)cot(x)

Answer 31

Now try.

Answer 32

Did you get it?